The chapter's techniques applied together. Every trigonometric equation is reduced to one of four basic forms, and the reduction is the whole skill: substituting for a compound argument and unwinding correctly, factoring a quadratic in disguise, using identities to match the functions or the arguments, converting a sum into a product so it can be set to zero factorwise, and collapsing a cosine plus a sine into a single sinusoid. Includes the rule that a trigonometric expression is never divided out, and how to count the solutions falling in a stated window.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.7 Trigonometric Equations and Inequalities, pp. 857-866
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 857-866 — the pages these objectives are drawn from
Warm-up
You can solve any basic equation: one function, one argument, one constant. Everything in this lesson is about getting an equation into that shape.
Discussion prompt
Solve sine of x equals one half. Now look at sine of 3x equals one half. What is the same and what is different?
Hint: Is the second one a different kind of equation, or the same kind with a different unknown?
Answer:
They are the same equation. The second has the form sine of something equals one half, and the something happens to be 3x rather than x. So the something is pi over six plus two pi k, or five pi over six plus two pi k.
\[ 3x = \tfrac{\pi}{6} + 2\pi k \;\Longrightarrow\; x = \tfrac{\pi}{18} + \tfrac{2\pi}{3}k \]
Then one extra step unwinds it. Notice that dividing by 3 divided the step as well, from two pi to two pi over three — which is the single most-missed detail in the whole lesson.
Concept
There are only four basic equation forms, one for each pair of functions, and each has a fixed recipe. Every other trigonometric equation is solved by transforming it into one of them.
So the question to ask of any equation is not how to solve it but what is stopping it from being basic — mismatched functions, mismatched arguments, a square, or a sum. Naming that obstacle picks the technique.
Figure (svg): The four basic equation forms and the recipe for each: cosine or sine solved in one revolution then stepped by two pi, secant or cosecant reciprocated first, tangent solved in one branch then stepped by pi, and cotangent converted to tangent
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 857-857
Section
Section 1
Concept
Cosine and sine are solved over one revolution and stepped by two pi. Tangent is solved over one branch and stepped by pi. Secant, cosecant and cotangent are converted to one of those first.
The neutral letter u is used for the argument on purpose. Whatever sits inside the function is what these recipes solve for, and it is only afterwards that the argument is unwound to reach x.
Figure (svg): The four basic equation forms and the recipe for each: cosine or sine solved in one revolution then stepped by two pi, secant or cosecant reciprocated first, tangent solved in one branch then stepped by pi, and cotangent converted to tangent
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 857-857
Picture it
Before choosing a method, name what is stopping the equation from being basic.
Figure (svg): Two columns pairing each obstacle in a trigonometric equation with the technique that removes it
Five obstacles and five techniques. Every worked example in this lesson is one of these rows, and knowing which row you are in is most of the battle.
Worked example
Example 10.7.1, part 3. The zero case has its own short answer.
\[ \text{Solve } \cot(3x) = 0 \text{ and list the solutions in } [0, 2\pi). \]
Recall the recipe for cotangent equal to zero
Why: The cotangent vanishes where the cosine does.
\[ u = \frac{\pi}{2} + \pi k \]
Set the argument equal to that
Why: The argument here is 3x.
\[ 3 x = \frac{\pi}{2} + \pi k \]
Divide by three, including the step
Why: Both the angle and the step are divided.
\[ x = \frac{\pi}{6} + (\frac{\pi}{3}) k \]
List the members in the window
Why: Six of them fit, since the step is pi over three and the window is two pi wide.
Figure (svg): The solution to Worked example a cotangent equation shown as a ladder of expressions, one row per legal move
\[ x = \tfrac{\pi}{6}, \tfrac{\pi}{2}, \tfrac{5\pi}{6}, \tfrac{7\pi}{6}, \tfrac{3\pi}{2}, \tfrac{11\pi}{6} \]
Verify: count them from the step
Why: The window has width two pi and the step is pi over three, so the number of members is two pi divided by pi over three, which is 6. That matches the list, and counting this way is a free check on whether any were missed.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 859-859
Sorting
Check the constant against that function's range before anything else.
Sort into buckets
Sort each equation.
Worked example
Checking the constant against the range comes before any work.
\[ \text{Solve } 3\cos(x) = 4 \text{ and } \sec(x) = \tfrac{1}{2}. \]
Isolate the function in the first
Why: Divide both sides by 3.
\[ \cos x = \frac{4}{3} \]
Check against the range
Why: Four thirds is about 1.333, outside the interval from negative one to one.
Examine the second
Why: A secant of one half is inside the gap in the secant's range.
\[ | \frac{1}{2} | < 1 \]
Conclude
Why: The secant never takes a value of absolute value under 1.
Figure (svg): The solution to Worked example an equation with no solutions shown as a ladder of expressions, one row per legal move
\[ 3\cos(x) = 4: \text{ none} \qquad \sec(x) = \tfrac{1}{2}: \text{ none} \]
Verify: state the reasons separately
Why: The first fails because a cosine cannot exceed 1; the second because a secant cannot be smaller than 1 in size. The two range conditions are opposites of each other, and checking the right one for the function in front of you is what makes this a one-line answer rather than a wasted page.
Trap
\[ \cos(2x) = -\tfrac{\sqrt{3}}{2} \;\Longrightarrow\; 2x = \tfrac{5\pi}{6} + 2\pi k \;\Longrightarrow\; x = \tfrac{5\pi}{12} + 2\pi k \]
Divide the angle by 2 and leave the step alone
Why: The step looks like a separate constant rather than part of the expression being divided.
But the whole right-hand side is divided by 2, and that includes the two pi k. Leaving it undivided loses half the solutions.
\[ x = \tfrac{5\pi}{12} + \pi k \]
Divide every term on the right by the coefficient
Why: Two pi k over 2 is pi k.
A quick check confirms it: the equation involves cosine of 2x, whose period is pi, so the solutions must repeat every pi rather than every two pi. If the step in your answer does not match the period of the compressed function, a division was missed.
Fill the middle
Solve sine of 4x equals root two over two.
Fill in the blanks
4x = \tfracpi/2pi/2 + 2\pi k \;\Longrightarrow\; x = \tfrac______ + ___k \qquad \text___ ___
Why: Dividing through by 4 divides the two pi k into pi over two times k. The check is that the period of sine of 4x is two pi over four, which is pi over two — and the step in the answer must equal that period. Whenever the two disagree, a division was missed.
Prediction
You are about to solve cosine of 5x equals 0.3 on the interval from zero to two pi.
Predict first
How many solutions do you expect?
Correct: 10.
Why: A cosine equation has two families. Each steps by two pi over five, the period of cosine of 5x, so each contributes five members to a window of width two pi. Two families of five gives ten solutions. Predicting the count before solving is a genuinely useful check: if your final list has a different length, something was dropped or duplicated.
Socratic
The strategy table is written for cosine of u rather than cosine of x.
Discussion prompt
Explain what that buys, and what would be lost by writing the recipes in terms of x.
Hint: What can the argument of a real equation actually be?
Answer:
The argument of a real equation is rarely just x. It might be 2x, or x over three minus pi, or 3x plus one. Writing the recipe for u makes it clear that the recipe solves for whatever sits inside the function, whatever that expression happens to be.
Writing it in terms of x would suggest the recipe applies only when the argument is x alone, and the almost inevitable consequence is applying it to the outside of the equation rather than the inside — the same error as setting a whole sinusoid equal to a quarter mark back in Lesson 10.5a. The neutral letter is a piece of notation doing real pedagogical work.
Section
Section 2
Concept
When the argument is a compound expression, treat it as a single unknown, apply the basic recipe to it, and only then solve for x. The unwinding is ordinary algebra and is where the errors live.
The book flags one error explicitly: after adding pi to both sides you may combine it with a pi over four, but never with the two pi k, because k is an arbitrary integer and the two terms are structurally different.
Figure (svg): A flow showing that an equation in cosine of two x is solved by treating two x as a single unknown u, solving for u, and only then dividing to find x
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 857-858
Picture it
Three stages, and the step changes at the last one.
Figure (svg): A flow showing that an equation in cosine of two x is solved by treating two x as a single unknown u, solving for u, and only then dividing to find x
The step halved because the argument had a coefficient of 2. That is not a special case; it happens whenever the argument is scaled.
Worked example
Example 10.7.1, part 1. Solve for 2x, then divide everything.
\[ \text{Solve } \cos(2x) = -\frac{\sqrt{3}}{2} \text{ and list the solutions in } [0, 2\pi). \]
Solve the basic equation for the argument
Why: A negative cosine gives quadrants two and three with reference angle pi over six.
\[ 2 x = 5 \pi / 6 + 2 \pi k\text{ or } 7 \pi / 6 + 2 \pi k \]
Divide every term by 2
Why: The step becomes pi k.
\[ x = 5 \pi / 12 + \pi k\text{ or } 7 \pi / 12 + \pi k \]
Substitute integer values of k
Why: Zero and one each give a member of both families inside the window.
\[ k = 0\text{ and } k = 1 \]
List the four solutions
Why: Two families, two members each.
\[ 5 \pi / 12, 7 \pi / 12, 17 \pi / 12, 19 \pi / 12 \]
Figure (svg): The solution to Worked example a coefficient in the argument shown as a ladder of expressions, one row per legal move
\[ x = \tfrac{5\pi}{12}, \; \tfrac{7\pi}{12}, \; \tfrac{17\pi}{12}, \; \tfrac{19\pi}{12} \]
Verify: check one solution
Why: At x equal to five pi over twelve, the argument is five pi over six, and the cosine there is negative root three over two. Correct. And four solutions in one revolution is what a doubled argument predicts, since the period is pi and each family contributes two.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 857-858
Faded example
Solve tangent of the quantity 2x minus pi over three, equals 1.
Fill in the blanks
2x - \frac7 pi/127 pi/24 = \frac______ + \pi k \;\Longrightarrow\; 2x = ___ + \pi k \;\Longrightarrow\; x = ___ + \frac______k
Why: Adding pi over three to pi over four gives seven pi over twelve, combining the two constants but leaving the pi k alone. Dividing everything by 2 halves both the constant and the step, giving seven pi over twenty-four with step pi over two — which matches the period of tangent of 2x, namely pi over two.
Worked example
Example 10.7.1, part 2. Two operations to unwind, in the right order.
\[ \text{Solve } \csc\left(\frac{1}{3}x - \pi\right) = \sqrt{2}. \]
Reciprocate to get a sine equation
Why: A cosecant of root two means a sine of root two over two.
\[ \sin u = \sqrt{2} / 2 \]
Solve for the argument
Why: Quadrants one and two, reference angle pi over four.
\[ u = \frac{\pi}{4} + 2 \pi k\text{ or } 3 \pi / 4 + 2 \pi k \]
Add pi to both sides
Why: The pi combines with the pi over four but NOT with the two pi k.
\[ \frac{x}{3} = 5 \pi / 4 + 2 \pi k \]
Multiply through by 3
Why: Every term, including the step.
\[ x = 15 \pi / 4 + 6 \pi k \]
Figure (svg): The solution to Worked example a coefficient and a constant shown as a ladder of expressions, one row per legal move
\[ x = \tfrac{15\pi}{4} + 6\pi k \quad\text{or}\quad x = \tfrac{21\pi}{4} + 6\pi k \]
Verify: check the step against the period
Why: The function is cosecant of x over three, whose period is two pi divided by one third, namely six pi. The step in the answer is six pi, matching. If the multiplication by 3 had been applied only to the first term, the step would have been two pi and the answer would have been wrong.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 858-858
Error analysis
A student unwinds an argument after solving the basic equation.
Annotate
On: \( \frac{x}{3} - \pi = \frac{\pi}{4} + 2\pi k \;\Longrightarrow\; \frac{x}{3} = \frac{\pi}{4} + 3\pi k \)
A term containing the arbitrary integer is structurally different from a constant. Keeping the family parameter visibly separate throughout the unwinding is the habit that prevents this.
Prediction
An equation involves sine of the quantity x over 4.
Predict first
What will the step be in the final answer?
Correct: 8 pi.
Why: Solving for the argument gives a step of two pi, and multiplying through by 4 to unwind multiplies the step by 4 as well, giving eight pi. That matches the period of sine of x over four, which is two pi divided by one quarter, namely eight pi. A stretched argument produces a longer period and therefore a larger step, which is the opposite of the compressed case.
Ranking
Solve an equation whose argument is a compound expression.
Put in order
Why: Converting first turns the equation into one of the forms the recipes cover. Solving for the whole argument comes before any unwinding, because the recipe applies to whatever is inside. Then the additions are undone before the multiplication, following the usual order for isolating a variable. The final check is free and catches the most common error.
Counterexample
A student proposes: the number of solutions in one revolution is always two for a cosine equation.
Discussion prompt
Give an equation that makes this false and explain what actually determines the count.
Hint: Try a coefficient inside the cosine.
Answer:
\[ \cos(4x) = \tfrac{1}{2} \text{ on } [0, 2\pi) \]
Each family steps by two pi over four, which is pi over two, so each contributes four members to a window of width two pi. Two families of four gives eight solutions, not two.
What determines the count is the period of the function as it appears in the equation. Two families, each contributing one member per period, and the number of periods in the window is two pi divided by the period. For cosine of n x that is n periods, so 2n solutions. The claim is true only when the coefficient is 1.
Section
Section 3
Concept
An equation containing a squared trigonometric function, or the same function to several powers, is a polynomial equation with an unusual variable. Naming that variable u makes the structure visible and the factoring routine.
The absolutely central rule: never divide both sides by a trigonometric expression. Doing so assumes it is nonzero, and the places where it is zero are usually solutions — so dividing silently discards them.
Figure (svg): An equation that is a quadratic in disguise, with the substitution u equals cosine of x making the quadratic structure visible before factoring
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 862-863
Picture it
The same equation, handled both ways, with different numbers of solutions found.
Figure (svg): A comparison showing that dividing both sides of an equation by a trigonometric expression loses solutions, while factoring keeps them
Dividing found two solutions and lost two. The lost ones are exactly the angles where the divided-out factor vanishes, and they are no less genuine than the others.
Worked example
Example 10.7.2, part 1. The book asks explicitly what goes wrong if you divide.
\[ \text{Solve } 3\sin^3(x) = \sin^2(x) \text{ and list the solutions in } [0, 2\pi). \]
Gather to one side
Why: Do not divide by sine squared.
\[ 3 \sin ^{3} x - \sin ^{2} x = 0 \]
Factor out the common factor
Why: Sine squared is common to both terms.
\[ \sin ^{2} x(3 \sin x - 1) = 0 \]
Set each factor to zero
Why: Two basic equations result.
\[ \sin x = 0\text{ or } \sin x = \frac{1}{3} \]
Solve each and collect
Why: The first gives multiples of pi; the second needs the arcsine.
Figure (svg): The solution to Worked example factor rather than divide shown as a ladder of expressions, one row per legal move
\[ x = 0, \; \pi, \; \arcsin\left(\tfrac{1}{3}\right), \; \pi - \arcsin\left(\tfrac{1}{3}\right) \]
Verify: see what dividing would have cost
Why: Dividing by sine squared would have left three sine x equal to 1 and found only the last two solutions. The values x equal to zero and pi genuinely satisfy the original equation, since both sides are zero there, and dividing by an expression that vanishes at exactly those points is what discards them.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 862-862
Prediction
You divide both sides of a trigonometric equation by cosine of x.
Predict first
What have you assumed?
Correct: That the cosine is never zero.
Why: Division by an expression is valid only where that expression is nonzero, so dividing by cosine of x silently assumes the cosine never vanishes. It vanishes twice per revolution, and those angles are frequently solutions of the original equation. The assumption is not merely unproven — it is false, which is why the operation loses solutions rather than requiring a caveat.
Worked example
Example 10.7.2, part 2. One identity turns it into a factorable quadratic.
\[ \text{Solve } \sec^2(x) = \tan(x) + 3. \]
Replace the secant squared
Why: The Pythagorean identity gives one plus tangent squared.
\[ 1 + \tan ^{2} x = \tan x + 3 \]
Gather to one side
Why: A quadratic in the tangent.
\[ \tan ^{2} x - \tan x - 2 = 0 \]
Substitute and factor
Why: Let u be the tangent.
\[ (u - 2) (u + 1) = 0 \]
Solve both basic equations
Why: One special value and one needing the arctangent.
\[ \tan x = 2\text{ or } \tan x = -1 \]
Figure (svg): The solution to Worked example a quadratic after an identity shown as a ladder of expressions, one row per legal move
\[ x = \arctan(2) + \pi k \quad\text{or}\quad x = -\tfrac{\pi}{4} + \pi k \]
Verify: list the window solutions
Why: On the interval from zero to two pi the first family gives arctangent of 2 and pi plus arctangent of 2; the second gives three pi over four and seven pi over four. Four solutions, which is what two tangent families contribute to a window of width two pi since each steps by pi.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 862-863
Trap
\[ \sin(2x) = \sqrt{3}\cos(x) \;\Longrightarrow\; 2\sin(x)\cos(x) = \sqrt{3}\cos(x) \]
Divide both sides by the cosine
Why: It appears on both sides and cancelling looks like an obvious simplification.
\[ 2\sin(x) = \sqrt{3} \;\Longrightarrow\; x = \tfrac{\pi}{3}, \tfrac{2\pi}{3} \]
Two solutions found. But the equation has four in one revolution, and the two missing ones are exactly where the cosine is zero.
\[ 2\sin(x)\cos(x) - \sqrt{3}\cos(x) = 0 \;\Longrightarrow\; \cos(x)\big(2\sin(x) - \sqrt{3}\big) = 0 \]
Move everything to one side and factor
Why: Now each factor can be set to zero in turn.
\[ x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{\pi}{3}, \tfrac{2\pi}{3} \]
Dividing by an expression is only valid when that expression is never zero, and a trigonometric function is zero somewhere in every revolution. The rule is therefore absolute in this setting: factor, never divide.
Faded example
Solve two cosine squared x plus cosine x minus one equals zero.
Fill in the blanks
2u^2 + u - 1 = 0 \;\Longrightarrow\; (2u - 1)(u + 1) = 0 \;\Longrightarrow\; \cos x = 1/2 \text___ \cos x = -1
Why: The quadratic factors as two u minus one times u plus one, giving u equal to one half or negative one. Substituting back gives two basic cosine equations, whose solutions are pi over three and five pi over three from the first, and pi from the second. Three solutions in one revolution, which is what a quadratic with two distinct roots gives when one of them is at a boundary value.
Sorting
Look for the same function appearing to more than one power.
Sort into buckets
Sort each equation.
Edge cases
Factoring the equation gave sine squared x times the quantity three sine x minus one, equal to zero.
Discussion prompt
The factor is sine SQUARED rather than sine. Does the squaring change the solution set, and does it change anything else?
Hint: When is a square zero?
Answer:
A square is zero exactly when the thing being squared is zero, so sine squared x equals zero gives the same solutions as sine x equals zero — the multiples of pi. The solution set is unchanged.
What does change is the multiplicity. Those roots are double roots, which in a graphical setting means the curve touches the axis there rather than crossing it. That distinction matters in calculus and when counting roots with multiplicity, but it makes no difference to a list of solutions.
So the practical answer is that the squaring can be ignored here, but noticing it is worthwhile: it is why the graph of the original equation flattens against the axis at those points rather than passing straight through.
Section
Section 4
Concept
If an equation contains two different functions, or the same function at two different arguments, no basic recipe applies until one of those mismatches is removed. The identity chapter is what removes them.
There is often more than one route. The book solves cosine of 3x equals cosine of 5x by sum-to-product rather than by expanding both into polynomials, because the resulting product is immediately factorable.
Figure (svg): Two columns pairing each obstacle in a trigonometric equation with the technique that removes it
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 863-865
Picture it
When a double angle identity is used, this is the shape the equation takes.
Figure (svg): An equation that is a quadratic in disguise, with the substitution u equals cosine of x making the quadratic structure visible before factoring
The identity's whole job was to make both terms involve the same function at the same argument. Once they do, ordinary algebra takes over.
Worked example
Example 10.7.2, part 3. A double angle identity turns it into a quadratic.
\[ \text{Solve } \cos(2x) = 3\cos(x) - 2. \]
Choose a form of the double angle identity
Why: The form with only cosines matches the other side.
\[ \cos 2 x = 2 \cos ^{2} x - 1 \]
Substitute and gather
Why: Everything to one side.
\[ 2 \cos ^{2} x - 3 \cos x + 1 = 0 \]
Factor as a quadratic
Why: Let u be the cosine.
\[ (2 u - 1) (u - 1) = 0 \]
Solve both basic equations
Why: Cosine equal to one half or to 1.
\[ \cos x = \frac{1}{2}\text{ or } \cos x = 1 \]
Figure (svg): The solution to Worked example matching the arguments shown as a ladder of expressions, one row per legal move
\[ x = \tfrac{\pi}{3}, \; \tfrac{5\pi}{3}, \; 0 \quad\text{on } [0, 2\pi) \]
Verify: check the boundary solution
Why: At x equal to zero the left side is the cosine of zero, which is 1, and the right side is three times 1 minus 2, which is also 1. It checks. Note that cosine equal to 1 gives a single family rather than two, since the line is tangent to the circle there.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 863-863
Matching
Name what is mismatched, then pick the identity that fixes it.
Match the pairs
Why: In each case the identity removes the specific mismatch. The cosine-only double angle form avoids introducing a sine; the Pythagorean identity turns the secant squared into tangents; sum-to-product converts a difference of cosines into a factorable product; and the sine double angle formula exposes a common factor of cosine x that can then be factored rather than divided out.
Worked example
Example 10.7.2, part 5. The zero on one side is the signal.
\[ \text{Solve } \cos(3x) = \cos(5x). \]
Move everything to one side
Why: The zero is what makes a product useful.
\[ \cos 5 x - \cos 3 x = 0 \]
Apply the sum-to-product identity
Why: Half-sum 4x, half-difference x, with the leading minus.
\[ -2 \sin(4 x) \sin(x) = 0 \]
Split into factors
Why: A product is zero when a factor is.
\[ \sin 4 x = 0\text{ or } \sin x = 0 \]
Solve and notice the containment
Why: The first gives multiples of pi over four; the second gives multiples of pi, which are already included.
\[ x = (\frac{\pi}{4}) k \]
Figure (svg): The solution to Worked example sum-to-product instead of expansion shown as a ladder of expressions, one row per legal move
\[ x = \tfrac{\pi}{4}k, \quad k \in \mathbb{Z} \]
Verify: check the containment claim
Why: The multiples of pi are pi k, and every one of those is a multiple of pi over four, namely pi over four times 4k. So the second solution set really is contained in the first and adds nothing. Noticing that avoids listing duplicates.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 864-864
Error analysis
A student solves cosine of 2x equals 3 cosine x minus 2, and picks a different double angle form.
Annotate
On: \( 1 - 2\sin^2(x) = 3\cos(x) - 2 \)
The three forms of the cosine double angle identity are not redundant. Choosing the one whose function already appears elsewhere in the equation is the whole reason there are three, and picking badly turns a one-step problem into a much longer one.
Fill the middle
Solve sine of 2x equals root three times cosine of x.
Fill in the blanks
2\sin x\cos x - \sqrtcos x\cos x = 0 \;\Longrightarrow\; sqrt3/2\big(2\sin x - \sqrt___\big) = 0 \;\Longrightarrow\; \cos x = 0 \text___ \sin x = ___
Why: The double angle identity turns sine of 2x into two sine x cosine x, exposing cosine x as a common factor. Factoring rather than dividing keeps both branches, giving four solutions on one revolution: pi over two and three pi over two from the first factor, and pi over three and two pi over three from the second.
Prediction
An equation reads sine x cosine of x over two, plus cosine x sine of x over two, equals 1.
Predict first
What should you notice first?
Correct: The left side is an expanded sum identity.
Why: Mixed terms of the form sine times cosine plus cosine times sine is exactly the expansion of the sine of a sum. Collapsing it gives the sine of x plus x over two, which is the sine of three x over two, and the equation becomes sine of three x over two equals 1 — a basic equation. Recognising an expanded identity is worth practising, because expanding further would make the problem much harder rather than easier.
Explain it to yourself
The book solved cosine of 3x equals cosine of 5x by sum-to-product rather than by expanding both sides as polynomials in cosine x.
Discussion prompt
Explain why that choice was better, and what the expansion route would have required.
Hint: What degree of polynomial would each expansion produce?
Answer:
Expanding cosine of 3x gives a cubic in cosine x and cosine of 5x gives a quintic. Setting them equal would leave a fifth-degree polynomial to factor, which is genuinely hard and has no general method.
Sum-to-product turns the same equation into a product of two sines, immediately factorable, and each factor is a basic equation. Two lines against a page of algebra.
The general lesson: a zero on one side is a signal to aim for a product, and sum-to-product is the tool that gets there. Reaching for expansion is the reflex, and it is usually the more expensive route when the two sides are the same function at different arguments.
Section
Section 5
Concept
An equation whose left side is a cosine plus a sine at the same argument is solved by rewriting that side as a single sinusoid. And once any equation is solved, listing the members inside a stated window is a separate small skill.
For a family with step two pi in a window of width two pi, exactly one member fits. Halving the step doubles the count, which is why compressed arguments produce so many more solutions.
Figure (svg): A number line from zero to two pi with the members of two solution families marked, showing how many of each family fall inside one revolution
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 865-866
Picture it
Two families with step pi in a window of width two pi give four solutions, not two.
Figure (svg): A number line from zero to two pi with the members of two solution families marked, showing how many of each family fall inside one revolution
Predicting the count from the step before listing anything is the cheapest possible check, and it catches both omissions and duplicates.
Worked example
Example 10.7.2, part 8. Nothing else works, and this does.
\[ \text{Solve } \cos(x) - \sqrt{3}\sin(x) = 2. \]
Notice both terms share an argument
Why: Both are functions of x, so the sum is a single sinusoid.
Find the amplitude
Why: Square and add the coefficients, then take the root.
\[ A = \sqrt{1 + 3} = 2 \]
Find the phase
Why: The coefficients give sine of phi equal to one half and cosine of phi negative root three over two.
\[ \phi = 5 \pi / 6 \]
Solve the resulting basic equation
Why: Two sine of the quantity x plus five pi over six equals 2, so the sine is 1.
\[ x = -\frac{\pi}{3} + 2 \pi k \]
Figure (svg): The solution to Worked example collapse a sum into a sinusoid shown as a ladder of expressions, one row per legal move
\[ x = -\tfrac{\pi}{3} + 2\pi k, \quad \text{giving } x = \tfrac{5\pi}{3} \text{ on } [0, 2\pi) \]
Verify: check the single solution
Why: At five pi over three the cosine is one half and the sine is negative root three over two. So the left side is one half minus root three times negative root three over two, which is one half plus three halves, namely 2. It checks. And a sine equal to 1 gives only one family, since the line is tangent, which is why only one solution appears.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 865-866
Fill the middle
How many solutions does a family with step pi over four have in the interval from zero to two pi?
Fill in the blanks
\textpi/4 = \frac8___} = ___
Why: Dividing the window width by the step gives the number of members, which is eight. Listing them means substituting k from zero through seven. If an equation produced two such families, sixteen solutions would be expected in one revolution — which sounds like a lot until you notice that a step of pi over four means the function has period pi over four and therefore completes eight cycles in the window.
Worked example
Example 10.7.1, part 5. The arctangent's range does the bounding.
\[ \text{Solve } \tan\left(\frac{x}{2}\right) = -3 \text{ and list the solutions in } [0, 2\pi). \]
Solve for the argument
Why: The arctangent handles the negative directly, with step pi.
\[ \frac{x}{2} = \arctan(-3) + \pi k \]
Unwind
Why: Multiply through by 2, doubling the step too.
\[ x = 2 \arctan(-3) + 2 \pi k \]
Bound the constant
Why: Since negative 3 is negative, the arctangent is between negative pi over two and zero, so twice it is between negative pi and zero.
\[ -\pi < 2 \arctan(-3) < 0 \]
Find which k fits
Why: k equal to zero gives a negative value; k equal to 1 shifts it into the window.
Figure (svg): The solution to Worked example list the solutions in a window shown as a ladder of expressions, one row per legal move
\[ x = 2\arctan(-3) + 2\pi \approx 4.7 \text{ on } [0, 2\pi) \]
Verify: check the count against the step
Why: The step is two pi and the window is two pi wide, so exactly one member fits — which matches. Bounding the constant analytically rather than reaching for a decimal is worth practising, since it is what an exam question asking for exact answers requires.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 860-861
Trap
\[ x = \tfrac{\pi}{12} + \tfrac{\pi}{3}k \;\text{ on } [0, 2\pi) \;\Longrightarrow\; x = \tfrac{\pi}{12}, \tfrac{5\pi}{12}, \tfrac{3\pi}{4} \]
Substitute a few values of k until the answers look plentiful enough
Why: Three solutions feels like a reasonable number and the pattern seems established.
But the step is pi over three and the window is two pi wide, so the number that fits is two pi divided by pi over three, which is six. Half the solutions are missing.
Compute the count first, then list exactly that many. Window width over step gives the number of members.
\[ \tfrac{2\pi}{\pi/3} = 6 \;\Longrightarrow\; k = 0, 1, 2, 3, 4, 5 \]
Substituting k from zero up to one less than the count generates them all in order, with no guessing and no duplicates. The same arithmetic works for any family, and it turns a hopeful search into a determinate one.
Prediction
An equation collapses to two sine of the quantity x plus phi, equals 2.
Predict first
How many solutions will it have in one revolution?
Correct: One.
Why: Dividing gives a sine equal to 1, which is the maximum value of the sine. The horizontal line at height 1 is tangent to the sine curve, touching it once per period rather than crossing it twice, so there is a single family with step two pi and exactly one member in a window of width two pi. Values of 1 and negative 1 are always the exceptional cases where the usual two families collapse into one.
Sorting
Count from the step, which comes from the coefficient inside the function.
Sort into buckets
Sort each equation by its number of solutions on the interval from zero to two pi.
Real world
A tidal model gives the depth at a harbour as 3.3 plus 1.9 times the cosine of 0.506 times the quantity t minus 3, with t in hours after midnight. A vessel needs at least 4.5 metres.
Discussion prompt
Set up the equation for when the depth is exactly 4.5 metres, describe how many solutions you expect in a 24 hour period, and say what the answer is used for.
Hint: The period is about 12.4 hours. How many periods fit in 24 hours?
Answer:
\[ 3.3 + 1.9\cos\big(0.506(t - 3)\big) = 4.5 \;\Longrightarrow\; \cos\big(0.506(t-3)\big) = \tfrac{1.2}{1.9} \approx 0.632 \]
Two families, each stepping by the period of about 12.4 hours. In a 24 hour window each family contributes about two members, so roughly four crossings per day — which is right, since the depth rises through 4.5 metres and falls back through it twice a day.
The four crossings bound two tidal windows during which the vessel can safely enter or leave. Publishing those windows is exactly what a tide table does, and the whole computation is one trigonometric equation solved on a stated interval. Predicting the count from the period first is what tells you not to stop after finding two.
Comparison
Fill the blanks from memory. Naming the obstacle chooses the method.
Comparison matrix
| Obstacle | Technique | Result |
|---|---|---|
| Compound argument | solve for the whole argument, then unwind | remember to divide the step too |
| A squared function | substitute u and factor | several basic equations |
| Different arguments | a double or multiple angle identity | one argument throughout |
| Different functions | a Pythagorean or quotient identity | one function throughout |
| A sum equal to zero | sum to product | a product that can be split |
| A cosine plus a sine | rewrite as one sinusoid | a basic equation |
Six rows, and every equation in this lesson is one of them. If none applies, the equation probably needs two of them in sequence.
Pattern
Whatever the equation looks like, the same five moves cover it.
Two checks are free and catch most errors: the step in the answer must equal the period of the function as it appears in the equation, and the number of window solutions must equal the window width divided by the step.
OpenStax Algebra and Trigonometry 2e, §9.5 Solving Trigonometric Equations §9.5
Check
A compound argument. Divide everything.
Check your understanding
Solve sine of 3x equals 1. What is the complete solution?
Answer: B
Why: A sine of 1 gives the single family 3x equal to pi over two plus two pi k. Dividing everything by 3 gives pi over six plus two pi over three times k. The step matches the period of sine of 3x, which is two pi over three.
Check
Factor, do not divide.
Check your understanding
How many solutions does 2 sine x cosine x equals cosine x have on the interval from zero to two pi?
Answer: C
Why: Moving everything to one side and factoring gives cosine x times the quantity two sine x minus one, equal to zero. The first factor gives pi over two and three pi over two; the second gives sine x equal to one half, so pi over six and five pi over six. Four solutions in total.
Check
Counting in a window. Use the step.
Check your understanding
A solution family is x equals pi over eight plus pi over two times k. How many members lie in the interval from zero to two pi?
Answer: B
Why: The count is the window width divided by the step, which is two pi divided by pi over two, namely 4. Substituting k equal to 0, 1, 2 and 3 gives pi over eight, five pi over eight, nine pi over eight and thirteen pi over eight, all inside the window, and k equal to 4 falls outside.
Real world
A Ferris wheel of radius 20 metres has its centre 22 metres up and completes a revolution every 40 seconds, starting a rider at the bottom. A photographer at a window 30 metres up wants to know when riders pass her eye level.
Discussion prompt
Model the height, set up the equation, and determine how many times per revolution and per hour a given rider passes 30 metres.
Hint: Height is 22 minus 20 cosine of the angle turned, since the rider starts at the bottom.
Answer:
\[ h(t) = 22 - 20\cos\left(\tfrac{\pi}{20}t\right) = 30 \;\Longrightarrow\; \cos\left(\tfrac{\pi}{20}t\right) = -\tfrac{8}{20} = -0.4 \]
\[ \tfrac{\pi}{20}t = \pm\arccos(-0.4) + 2\pi k \;\Longrightarrow\; t = \pm\tfrac{20}{\pi}\arccos(-0.4) + 40k \]
Two families, each stepping by 40 seconds — one revolution. So the rider passes 30 metres twice per revolution, once going up and once coming down, which is the physically obvious answer arrived at algebraically.
At 40 seconds per revolution there are 90 revolutions per hour, so 180 passes per hour. The count came from the step, exactly as in the window-counting technique: the number of solutions in an interval is the interval divided by the step, and here the interval is an hour and the step is 40 seconds.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
May you divide both sides of sine x cosine x equals cosine x by cosine x?
Correct: No, it discards the solutions where the cosine is zero.
\[ \cos(x)\big(\sin(x) - 1\big) = 0 \;\Longrightarrow\; \cos x = 0 \;\text{ or }\; \sin x = 1 \]
Why: Dividing by an expression is valid only where that expression is nonzero, and the cosine vanishes twice per revolution — at exactly the angles that satisfy the original equation, since both sides become zero there. So the division loses genuine solutions. The second option is closer to defensible: dividing and then separately checking the discarded case does work, but it is strictly more effort than factoring and is easier to forget. The fourth option overstates it; division by a nonzero constant is entirely fine.
Explain it
They are staring at cos 2x equals 3 cos x minus 2 and say they have never seen an equation like it.
Discussion prompt
In no more than five sentences, show them that they have. Give them the question to ask of any trigonometric equation.
Hint: What is stopping it from being an equation they can already do?
Answer:
A usable answer: the only problem is that one term says 2x and the other says x. Use the identity that turns cosine of 2x into two cosine squared x minus one, and now everything is in cosine x. Call cosine x by the name u and look at what you have: two u squared minus three u plus one equals zero — an ordinary quadratic you could have factored last year.
The question to ask of any trigonometric equation is what is stopping this from being one function, one argument, equal to a number? Whatever the answer is, there is an identity or a factoring step that removes it, and afterwards you are back on ground you already know.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Unwinding is fixed by checking the final step against the period of the function as it appears in the equation. Factoring is fixed by an absolute rule: everything to one side, never divide. Identity choice is fixed by naming the obstacle out loud before reaching for anything. Window listing is fixed by computing the count as window width over step before substituting any k. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page write the four basic forms and the recipe for each, noting beside two of them which constants give no solutions. Below that, make a two-column table of obstacles and techniques with six rows, covering compound arguments, squares, mismatched arguments, mismatched functions, sums equal to zero, and a cosine plus a sine. In the middle left, solve cosine of 2x equals negative root three over two completely, showing the unwinding and marking clearly where the step gets divided. In the middle right, solve sine of 2x equals root three cosine x by factoring, and write underneath which two solutions dividing by cosine x would have lost. In the bottom left, write the formula for how many members of a family fall in a window, and use it on a family with step pi over three. In the bottom right, collapse cosine x minus root three sine x into a single sinusoid and solve it equal to 2. Finally, circle the single rule on the page that, if broken, silently loses solutions rather than giving a visibly wrong answer.
The circled rule should be never dividing both sides by a trigonometric expression. It is the only one whose violation produces an answer that looks complete and is not, which is what makes it the most dangerous.
Recap
Five things, and the second is the rule that silently costs marks if it is broken.
| If the question says | Your first move is |
|---|---|
| Solve this equation | Check the constant against the range |
| The same function appears twice | Substitute u and factor |
| A function appears on both sides | Move everything over and aim for a product |
| The arguments differ | Use a double or multiple angle identity |
| List the solutions in an interval | Compute the count as width over step first |
One lesson of Chapter 10 remains. Inequalities use the same reductions but ask a different question: not where the two sides are equal, but where one exceeds the other — and the answer is a set of intervals rather than a set of points.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 857-866 — everything on these slides traces back here
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