General solutions of trigonometric equations, solutions restricted to an interval, solving by factoring and by the quadratic formula, discarding impossible roots, and recognising extraneous solutions created by squaring.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 14 — Trigonometric Graphs, Identities, and Equations
Solve Trigonometric Equations
Objectives
Five outcomes. Not proving it, but finding where it is true.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 931-935 — the lesson these objectives are drawn from
Warm-up
Lesson 14.3 verified identities, which hold for every angle.
Discussion prompt
Now take sine x equal to root 3 over 2. Is that true for every x? If not, for which x is it true?
Hint: Where does a horizontal line cross the sine graph?
Answer:
Not for every x. It holds at pi over 3 and at 2 pi over 3, and then again a full turn later, and a turn after that, forever.
\[ x = \tfrac{\pi}{3}+2n\pi \quad \text{or} \quad x = \tfrac{2\pi}{3}+2n\pi \]
Finding those values is solving, and because the graph repeats there are always infinitely many. The formula above, with n any integer, is called the general solution.
Concept
A trigonometric equation is true only for particular values. Because the functions repeat, the solution set is infinite, and it is described by a general solution: the solutions in one period plus a multiple of that period.
general solution — A formula describing every solution of a trigonometric equation, written as the solutions within one period plus an integer multiple of the period. For sine and cosine that multiple is 2 n pi; for tangent it is n pi.
\[ x = \tfrac{\pi}{3}+2n\pi \;\text{ or }\; x = \tfrac{2\pi}{3}+2n\pi \]
The inverse functions return only one solution each, so the others must always be built by hand from the symmetry of the graph.
Figure (svg): Two columns comparing an identity with an equation to solve
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 931-931
Section
Section 1
Concept
Isolate the trigonometric function, use an inverse to find one solution, use the graph's symmetry to find the other within a period, then add a multiple of the period to describe every solution.
\[ x = x_1+2n\pi \; \text{ or } \; x = x_2+2n\pi \]
For sine, the second solution in a turn is pi minus the first; for cosine it is 2 pi minus it; for tangent there is only one per period of pi.
Figure (svg): Two solutions per turn, repeating forever
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 931-931 — Solve a trigonometric equation
Picture it
Example 1: sine x equal to root 3 over 2.
Figure (svg): Two solutions per turn, repeating forever
The horizontal line cuts the circle twice, giving pi over 3 and 2 pi over 3, and every full turn reproduces both.
Worked example
Example 1.
\[ \text{Solve } 2\sin x-\sqrt3 = 0. \]
Isolate the sine
Why: Add root 3, then divide by 2.
\[ \sin x = \sqrt{3} / 2 \]
Find one solution
Why: The inverse sine of root 3 over 2.
\[ x = \frac{\pi}{3} \]
Find the other in the same turn
Why: Pi minus the first.
\[ x = 2 \pi / 3 \]
Add multiples of the period
Why: Sine repeats every 2 pi.
\[ +2 n \pi\text{ to each} \]
Figure (svg): Two solutions per turn, repeating forever
\[ x = \tfrac{\pi}{3}+2n\pi \;\text{ or }\; x = \tfrac{2\pi}{3}+2n\pi \]
Verify: check both by graphing
Why: The line y equal to root 3 over 2 meets the sine curve at infinitely many points, and they occur in pairs spaced 2 pi apart. Graphing both sides is the standard visual check, and it makes the infinitude obvious.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 931-931
Fill the middle
Example 1.
Fill in the blanks
\sin x = \tfrac2___: \; x_1 = \tfrac______, \; x_2 = \pi-\tfrac______ = \tfrac___\pi}___
Why: Two pi over 3. For a sine equation the second solution in a turn is always pi minus the first, because sine is symmetric about pi over 2.
Worked example
Guided Practice 1.
\[ \text{Find the general solution of } 2\sin x+4 = 5. \]
Isolate the sine
Why: Subtract 4, then divide by 2.
\[ \sin x = \frac{1}{2} \]
Find one solution
Why: The inverse sine of one half.
\[ x = \frac{\pi}{6} \]
Find the other
Why: Pi minus pi over 6.
\[ x = 5 \pi / 6 \]
Write the general solution
Why: Add 2n pi to each.
\[ \frac{\pi}{6} + 2 n \pi\text{ or } 5 \pi / 6 + 2 n \pi \]
Figure (svg): The solution to Worked example another general solution shown as a ladder of expressions, one row per algebraic move
\[ x = \tfrac{\pi}{6}+2n\pi \;\text{ or }\; x = \tfrac{5\pi}{6}+2n\pi \]
Verify: check the second solution's sine
Why: Sine of 5 pi over 6 is one half, since 5 pi over 6 has reference angle pi over 6 and lies in quadrant two where sine is positive. That is the whole reason the second solution is pi minus the first rather than anything else.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 933-933
Trap
\[ \sin x = \tfrac{\sqrt3}{2} \]
Take the inverse sine and stop
Why: The calculator gives one answer.
\[ x = \tfrac{\pi}{3} \quad \text{(incomplete)} \]
Two pi over 3 also has sine root 3 over 2, and so do infinitely many other angles. One answer out of infinitely many is not a solution set.
\[ x = \tfrac{\pi}{3} \text{ and } x = \pi-\tfrac{\pi}{3} = \tfrac{2\pi}{3} \]
Use the graph's symmetry for the second, then add periods
Why: The inverse returns only its range's value.
\[ x = \tfrac{\pi}{3}+2n\pi \;\text{ or }\; x = \tfrac{2\pi}{3}+2n\pi \]
This is the same restriction that made the inverse functions well defined in Lesson 13.4, now costing you the other solutions.
Matching
How much to add for the general solution.
Match the pairs
Why: Tangent is the exception because its period is pi rather than 2 pi, so one solution per period already covers everything. Sine and cosine both need two, but they are found differently: pi minus for sine, 2 pi minus for cosine.
Sorting
Between 0 and 2 pi.
Sort into buckets
Sort each equation.
The extremes are the exception: sine equal to 1 or negative 1 has only one solution per turn, because those are the turning points where the graph does not cross but touches.
Prediction
Commit before reasoning.
Predict first
Why does a solvable trigonometric equation always have infinitely many solutions?
Correct: Because the functions are periodic, so any solution plus a whole period is another solution.
\[ \sin(x+2\pi) = \sin x \]
Why: If sine x equals one half then sine of x plus 2 pi equals one half too, since adding a full turn changes nothing. So one solution generates infinitely many at once. That is exactly why the answer is written as a formula with an integer n in it rather than as a list, and why every problem either asks for the general solution or restricts attention to an interval.
Section
Section 2
Concept
When an interval is specified, find the general solution first and then substitute integer values of n, keeping only the results that land inside the interval.
\[ x \approx 0.322+n\pi \;\text{ or }\; x \approx -0.322+n\pi \]
For a tangent equation the period is pi, so four solutions can appear in a single interval of length 2 pi.
Figure (svg): Four solutions of a tangent equation inside one turn
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 932-932 — Solve a trigonometric equation in an interval
Picture it
Example 2: nine tangent squared x plus 2 equals 3.
Figure (svg): Four solutions of a tangent equation inside one turn
Two values of tangent, each repeating every pi, produce four angles between 0 and 2 pi.
Worked example
Example 2.
\[ \text{Solve } 9\tan^2x+2 = 3 \text{ on } 0 \le x < 2\pi. \]
Isolate the squared function
Why: Subtract 2, divide by 9.
\[ \tan ^{2} x = \frac{1}{9} \]
Take square roots
Why: Both signs.
\[ \tan x = \frac{1}{3}\text{ or } -\frac{1}{3} \]
Write the general solutions
Why: Tangent's period is pi.
\[ 0.322 + n \pi\text{ and } -0.322 + n \pi \]
Filter for the interval
Why: Choose the n that land in [0, 2 pi).
\[ 0.322, 2.820, 3.464, 5.961 \]
Figure (svg): Four solutions of a tangent equation inside one turn
\[ 0.322, \; 2.820, \; 3.464, \; 5.961 \]
Verify: check the spacing between them
Why: The four values differ from one another by about 2.498 and 0.644 alternately, and each pair from the same branch is exactly pi apart: 3.464 minus 0.322 is 3.142, and 5.961 minus 2.820 is also 3.142. That spacing is the strongest single check on a tangent solution set.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 932-932
Fill the middle
Example 2.
Fill in the blanks
\tan^2x = \tfrac3___ \;\Longrightarrow\; \tan x = \pm\tfrac______}
Why: Plus or minus one third. Both signs must be carried, since squaring either one gives the same result.
Worked example
Guided Practice 2.
\[ \text{Solve } 3\csc^2x = 4 \text{ on } 0 \le x < 2\pi. \]
Isolate and invert
Why: Cosecant squared is 4/3, so sine squared is 3/4.
\[ \sin ^{2} x = \frac{3}{4} \]
Take square roots
Why: Both signs.
\[ \sin x = \sqrt{3} / 2\text{ or } -\sqrt{3} / 2 \]
Solve the positive case
Why: Quadrants one and two.
\[ \frac{\pi}{3}\text{ and } 2 \pi / 3 \]
Solve the negative case
Why: Quadrants three and four.
\[ 4 \pi / 3\text{ and } 5 \pi / 3 \]
Figure (svg): The solution to Worked example a cosecant equation shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{\pi}{3}, \; \tfrac{2\pi}{3}, \; \tfrac{4\pi}{3}, \; \tfrac{5\pi}{3} \]
Verify: check all four have the same reference angle
Why: Every one of the four has reference angle pi over 3, and they occupy all four quadrants — which is what taking a square root of a squared function does. Squaring destroyed both the sign of sine and the quadrant, so all four possibilities return.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 933-933
Error analysis
A student solves 9 tan squared x plus 2 equals 3.
Annotate
On: \( \tan^2x = \tfrac{1}{9} \;\Longrightarrow\; \tan x = \tfrac{1}{3} \)
Every square root in an equation carries a plus-or-minus. Forgetting it here halves the answer set, and the loss is silent.
Fill the middle
Example 2.
Fill in the blanks
-0.322+\pi \approx 2.820 \text___ [0,2\pi)
Why: About 2.820. Taking n equal to 1 on the negative branch moves a solution that was outside the interval into it.
Ranking
From Example 2, smallest first.
Put in order
Why: The values are 0.322, 2.820, 3.464, 5.961 and about 6.283. All four solutions fall just inside the interval, and the interval's own endpoint is excluded.
Prediction
Commit before reasoning.
Predict first
A sine equation typically has two solutions in a turn. Why did this tangent equation have four?
Correct: Because tangent's period is pi, so each of the two tangent values recurs twice inside 2 pi.
\[ 2 \text{ values} \times 2 \text{ periods of } \pi \text{ in } 2\pi = 4 \]
Why: Squaring gave two values of tangent, positive and negative one third, and each of those has one solution per period of pi. Two periods fit inside 2 pi, so each value contributes two angles, making four in all. Had the equation been a sine equation squared instead, the same reasoning would also give four, but for a different reason: two values of sine with two solutions each per turn.
Section
Section 3
Concept
Move everything to one side, factor, and set each factor equal to zero. Then check each resulting equation against the function's range, since some may be impossible.
\[ \sin x(\sin x+3)(\sin x-3) = 0 \]
Sine and cosine never leave the band from negative one to one, so equations demanding a value outside it contribute nothing.
Figure (svg): A trigonometric equation solved by factoring
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 933-933 — Solve by factoring
Picture it
Example 4: sine cubed x minus 9 sine x equals zero.
Figure (svg): A trigonometric equation solved by factoring
Two of the three factors demand a sine of plus or minus 3, which no angle has. Only the first factor produces solutions.
Worked example
Example 4.
\[ \text{Find the general solution of } \sin^3x-9\sin x = 0. \]
Factor out the common factor
Why: Sine x is in both terms.
\[ \sin x(\sin ^{2} x - 9) = 0 \]
Factor the difference of squares
Why: Nine is 3 squared.
\[ \sin x(\sin x + 3) (\sin x - 3) = 0 \]
Set each factor to zero
Why: Three equations.
\[ \sin x = 0, -3, 3 \]
Discard the impossible ones
Why: Sine cannot exceed 1 in size.
\[ \text{only } \sin x = 0\text{ survives} \]
Figure (svg): A trigonometric equation solved by factoring
\[ x = 2n\pi \;\text{ or }\; x = \pi+2n\pi \]
Verify: notice the two branches combine
Why: Two n pi and pi plus 2n pi together are exactly the integer multiples of pi, so the answer could be written more compactly as x equals n pi. Both forms are correct, and the compact one is worth spotting.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 933-933
Fill the middle
Example 4.
Fill in the blanks
\sin^2x-9 = (\sin x+3)(\sin x-3)
Why: Three. Neither resulting factor can ever be zero, since sine stays between negative 1 and 1.
Worked example
Guided Practice 4 and 7.
\[ \text{Solve } \sin^3x-\sin x = 0 \text{ generally, and } \tan^2x-\sin x\tan^2x = 0 \text{ on } 0 \le x \le \pi. \]
First: factor
Why: Sine times sine squared minus 1.
\[ \sin x(\sin x - 1) (\sin x + 1) = 0 \]
First: solve each
Why: Sine equal to 0, 1 or -1.
\[ \text{every multiple of } \frac{\pi}{2} \]
Second: factor
Why: Tangent squared times 1 minus sine.
\[ \tan ^{2} x(1 - \sin x) = 0 \]
Second: solve and check the domain
Why: Sine equal to 1 gives pi/2, where tangent is undefined.
\[ x = 0\text{ and } x = \pi\text{ only} \]
Figure (svg): The solution to Worked example two more by factoring shown as a ladder of expressions, one row per algebraic move
\[ x = \tfrac{n\pi}{2}; \qquad x = 0, \; \pi \]
Verify: check why pi over 2 is rejected
Why: At x equal to pi over 2 the tangent is undefined, so the original expression has no value there and cannot be zero. A factored form can suggest a solution that the original equation's domain forbids, which is a second reason to check every candidate in the ORIGINAL equation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 934-934
Trap
\[ \sin^3x-9\sin x = 0 \]
Divide both sides by sine x
Why: It appears in both terms.
\[ \sin^2x-9 = 0 \;\Longrightarrow\; \text{no solutions} \quad \text{(wrong)} \]
Dividing by sine x throws away every solution where sine x is zero — which here is the only place solutions exist.
\[ \sin x(\sin^2x-9) = 0 \]
Factor rather than divide
Why: Then the zero-product property keeps every case.
\[ \sin x = 0 \;\Longrightarrow\; x = n\pi \]
Dividing by an expression that might be zero is illegal, and in trigonometric equations that expression is usually zero somewhere.
Sorting
Check against the function's range.
Sort into buckets
Sort each equation.
Three of five contribute nothing at all. Checking each factor's equation against the range before reaching for a calculator saves both time and error messages.
Matching
Factor first.
Match the pairs
Why: The last needed rewriting rather than factoring: two sine x equals one over sine x gives sine squared x equal to one half. The third looked as though pi over 2 should be a solution but the domain forbade it.
Prediction
Commit before reasoning.
Predict first
Dividing sine cubed x minus 9 sine x equals 0 by sine x looks tidy. Why is it wrong?
Correct: Because sine x can be zero, and dividing by it discards exactly the solutions the equation has.
\[ \text{divide by } \sin x \;\Longrightarrow\; \text{lose } \sin x = 0 \]
Why: Division is legal only by a nonzero quantity, and here sine x is zero precisely at the solutions. Dividing therefore removes all of them and leaves an equation with none. This is the same rule that forbade dividing by x when solving x cubed equals 4x in Chapter 5, and it bites harder in trigonometry because trigonometric expressions are zero so often.
Section
Section 4
Concept
An equation of the form a times the function squared, plus b times the function, plus c equals zero can be solved by factoring or by the quadratic formula, treating the function as a single unknown.
\[ \cos x = \frac{5\pm\sqrt{17}}{2} \]
Each root then has to be checked against the function's range, and only the acceptable ones lead to angles.
Figure (svg): A trigonometric equation solved by the quadratic formula
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 933-933 — Use the quadratic formula
Picture it
Example 5: cosine squared x minus 5 cosine x plus 2 equals zero.
Figure (svg): A trigonometric equation solved by the quadratic formula
The formula returns 4.56 and 0.44. Only the second can be a cosine, and it gives a single angle in the stated interval.
Worked example
Example 5.
\[ \text{Solve } \cos^2x-5\cos x+2 = 0 \text{ on } 0 \le x \le \pi. \]
Recognise the form
Why: With cosine x as the unknown.
\[ a = 1, b = -5, c = 2 \]
Apply the quadratic formula
Why: Five plus or minus the root of 17, over 2.
\[ \cos x = 4.56\text{ or } 0.44 \]
Reject the impossible root
Why: Cosine never exceeds 1.
\[ 4.56\text{ discarded} \]
Take the inverse cosine
Why: Of 0.44, in the stated interval.
\[ x =\text{ about } 1.12 \]
Figure (svg): A trigonometric equation solved by the quadratic formula
\[ x \approx 1.12 \]
Verify: check the answer in the original equation
Why: Cosine of 1.12 is about 0.435, and 0.435 squared minus 5 times 0.435 plus 2 is about 0.189 minus 2.175 plus 2, which is about 0.014 — zero to the accuracy of the rounding. Substituting back is the only way to catch a formula slip.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 933-933
Fill the middle
Example 5.
Fill in the blanks
\cos x = \frac17}___ = \frac___}}}___
Why: Seventeen. The two roots are about 4.56 and 0.44, and only the smaller can be a cosine.
Worked example
Guided Practice 6.
\[ \text{Solve } 2\sin x = \csc x \text{ on } 0 \le x \le \pi. \]
Replace the cosecant
Why: One over sine x.
\[ 2 \sin x = 1 / \sin x \]
Clear the fraction
Why: Multiply by sine x.
\[ 2 \sin ^{2} x = 1 \]
Solve for sine
Why: Sine squared is one half.
\[ \sin x =\text{ plus or minus } \sqrt{2} / 2 \]
Keep what the interval allows
Why: Sine is not negative on [0, pi].
\[ x = \frac{\pi}{4}\text{ and } 3 \pi / 4 \]
Figure (svg): The solution to Worked example a quadratic solved by rearranging shown as a ladder of expressions, one row per algebraic move
\[ x = \tfrac{\pi}{4}, \; \tfrac{3\pi}{4} \]
Verify: check that multiplying by sine x was safe
Why: Sine x is zero at 0 and at pi, and at those angles the original equation's cosecant is undefined anyway, so nothing was lost. When clearing a fraction, always ask whether the multiplier could be zero and whether the original was even defined there.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 934-934
Error analysis
A student solves cosine squared x minus 5 cosine x plus 2 equals zero.
Annotate
On: \( \cos x \approx 4.56 \;\Longrightarrow\; x = \cos^{-1}4.56 \)
A calculator error message here is not a mistake to fix but information: the root is outside the range and contributes no solutions.
Fill the middle
Guided Practice 6.
Fill in the blanks
2\sin x = \frac1___ \;\Longrightarrow\; 2\sin^2x = ___
Why: One. Sine squared x is then one half, so sine x is plus or minus root 2 over 2.
Comparison
Fill the blanks. Two routes to a quadratic form.
Comparison matrix
| Question | Factoring | Quadratic formula |
|---|---|---|
| When to use | when the factors are visible | when they are not |
| Example | sin^3 x - 9 sin x = 0 | cos^2 x - 5 cos x + 2 = 0 |
| Roots to check | yes, against the range | yes, against the range |
| Risk | dividing instead of factoring | taking an impossible inverse |
Both routes end at the same place: one or more values of the function, each of which then needs checking against the range before any inverse is taken.
Prediction
Commit before reasoning.
Predict first
The quadratic formula returned cosine x equal to 4.56. Why is that not an error in the algebra?
Correct: Because the algebra treated cosine x as an ordinary unknown, and nothing in it knows the range is -1 to 1.
\[ -1 \le \cos x \le 1 \;\Longrightarrow\; 4.56 \text{ rejected} \]
Why: The quadratic formula solves for a number, not for a cosine, so it happily returns values no cosine could take. Supplying that extra knowledge is the solver's job, not the formula's. The same thing happens whenever a substitution hides a restriction, which is why every quadratic-form problem in this lesson ends with a range check before an inverse is taken.
Section
Section 5
Concept
Squaring both sides of an equation can introduce solutions that satisfy the squared equation but not the original one. Every candidate must therefore be checked in the ORIGINAL equation.
\[ 1+\cos x = \sin x \]
Squaring destroys sign information, so a candidate satisfying the squared version may have the two sides equal in size but opposite in sign.
Figure (svg): Two curves meeting at only two of three candidate points
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 934-934 — Solve an equation with an extraneous solution
Picture it
Example 6: one plus cosine x equals sine x.
Figure (svg): Two curves meeting at only two of three candidate points
The two curves meet at pi over 2 and pi only. The third candidate, 3 pi over 2, has the sides equal in size but opposite in sign.
Worked example
Example 6.
\[ \text{Solve } 1+\cos x = \sin x \text{ on } 0 \le x < 2\pi. \]
Square both sides
Why: Then expand the left.
\[ 1 + 2 \cos x + \cos ^{2} x = \sin ^{2} x \]
Use the Pythagorean identity
Why: Sine squared is 1 minus cosine squared.
\[ 2 \cos ^{2} x + 2 \cos x = 0 \]
Factor and solve
Why: Two cosine x times cosine x plus 1.
\[ \cos x = 0\text{ or } -1 \]
Check all three candidates
Why: Pi over 2, pi, and 3 pi over 2.
\[ 3 \pi / 2\text{ fails} \]
Figure (svg): Two curves meeting at only two of three candidate points
\[ x = \tfrac{\pi}{2}, \; \pi \]
Verify: see why 3 pi over 2 failed
Why: There the left side is 1 plus 0, which is 1, and the right side is negative 1. Squaring made both sides 1 and hid the difference in sign. That is precisely the information squaring destroys, and why the check cannot be skipped.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 934-934
Fill the middle
Example 6.
Fill in the blanks
x = \tfrac-1___: \; 1+0 = 1 \text___ \sin\tfrac______ = ___
Why: Negative 1. The two sides are 1 and negative 1, equal in size but opposite in sign, which is exactly what squaring hides.
Worked example
Guided Practice 5.
\[ \text{Find the general solution of } 1-\cos x = \sqrt3\sin x. \]
Square both sides
Why: Three sine squared on the right.
\[ 1 - 2 \cos x + \cos ^{2} x = 3 - 3 \cos ^{2} x \]
Collect and simplify
Why: Divide through by 2.
\[ 2 \cos ^{2} x - \cos x - 1 = 0 \]
Factor and solve
Why: Two cosine x plus 1, times cosine x minus 1.
\[ \cos x = -\frac{1}{2}\text{ or } 1 \]
Check the candidates
Why: Zero, 2 pi/3 and 4 pi/3.
\[ 4 \pi / 3\text{ fails} \]
Figure (svg): The solution to Worked example another squaring problem shown as a ladder of expressions, one row per algebraic move
\[ x = 2n\pi \;\text{ or }\; x = \tfrac{2\pi}{3}+2n\pi \]
Verify: check the rejected candidate
Why: At 4 pi over 3 the left side is 1 plus one half, which is three halves, while the right is root 3 times negative root 3 over 2, which is negative three halves. Equal in size, opposite in sign — the signature of an extraneous root from squaring.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 934-934
Trap
\[ 1+\cos x = \sin x \]
Square, solve, and report every candidate
Why: The algebra was all legal, so the answers are trusted.
\[ x = \tfrac{\pi}{2}, \; \pi, \; \tfrac{3\pi}{2} \quad \text{(one is false)} \]
Squaring is not reversible. Every step was legal, but the squared equation has more solutions than the original.
\[ \text{substitute each candidate into the ORIGINAL} \]
Check every candidate before reporting
Why: Only in the original equation, never the squared one.
\[ x = \tfrac{\pi}{2}, \; \pi \]
Graphing both sides confirms it: the two curves cross exactly twice on the interval.
Sorting
Which operations are not reversible.
Sort into buckets
Sort each step.
Multiplying by an expression that could be zero carries the same risk as squaring, for the same reason: it is not an operation you can simply undo.
Matching
Substitute into the original equation.
Match the pairs
Why: Both rejected candidates have the two sides equal in size and opposite in sign, which is the unmistakable signature of a root created by squaring. A candidate that fails for any other reason usually indicates an arithmetic slip instead.
Prediction
Commit before reasoning.
Predict first
Why can squaring both sides produce solutions the original equation does not have?
Correct: Because a equals b and a equals negative b both square to the same equation, so the squared version cannot tell them apart.
\[ a = -b \;\Longrightarrow\; a^2 = b^2 \text{ too} \]
Why: The squared equation is satisfied by every solution of the original AND by every solution of the sign-flipped version, so its solution set is strictly larger. The candidates that came from the wrong sign are the extraneous ones. This is the same phenomenon you met solving radical equations in Chapter 6, and the remedy is identical: check every candidate in the original.
Comparison
Fill the blanks. Match the method to the shape.
Comparison matrix
| Shape of the equation | Method | Extra care needed |
|---|---|---|
| One function, isolated | inverse, then symmetry, then periods | the inverse gives only one solution |
| A squared function | take both square roots | do not lose the negative root |
| A common factor | factor and use zero product | never divide by the factor |
| Quadratic in the function | factor or use the formula | check each root against the range |
Whatever route you take, the equation ends as one or more statements of the form function equals number, and from there the general solution follows the same way every time.
Pattern
Reduce, solve, check.
Factor rather than divide. Dividing by a trigonometric expression discards the solutions where it is zero.
OpenStax Algebra and Trigonometry 2e, §9.5 Solving Trigonometric Equations §9.5
Check
The inverse gives only one of two.
Check your understanding
What is the general solution of 2 sin x - root 3 = 0?
Answer: A
Why: sin x = root3/2 gives pi/3 and pi - pi/3, each repeating every 2 pi.
Check
Check every factor against the range.
Check your understanding
What is the general solution of sin cubed x - 9 sin x = 0?
Answer: A
Why: Factoring gives sin x = 0, 3 or -3; only sin x = 0 is possible.
Check
Squaring demands a check.
Check your understanding
Solving 1 + cos x = sin x on 0 to 2 pi gives candidates pi/2, pi and 3 pi/2. Which are actual solutions?
Answer: A
Why: At 3 pi/2 the left side is 1 and the right side is -1, so it fails.
Real world
The water depth in the Bay of Fundy is modelled by d equal to 35 minus 28 cosine of pi t over 6.2, with d in feet and t in hours after midnight.
Discussion prompt
At what times in a day is the depth 7 feet, and at what times is it 63 feet?
Hint: Substitute the depth and solve for t.
Answer:
\[ 35-28\cos\tfrac{\pi t}{6.2} = 7 \;\Longrightarrow\; \cos\tfrac{\pi t}{6.2} = 1 \;\Longrightarrow\; \tfrac{\pi t}{6.2} = 2n\pi \]
\[ t = 12.4n \;\Longrightarrow\; t = 0 \text{ and } t = 12.4 \text{ (midnight and 12:24 p.m.)} \]
\[ 35-28\cos\tfrac{\pi t}{6.2} = 63 \;\Longrightarrow\; \cos\tfrac{\pi t}{6.2} = -1 \;\Longrightarrow\; t = 6.2+12.4n \]
The depth is 7 feet at midnight and 12:24 p.m., and 63 feet at 6:12 a.m. and 6:36 p.m.
Notice that these are the two extreme depths, which is why each gave a cosine of exactly plus or minus 1 and so only one solution per period rather than two. The 12.4 hour period is the real tidal cycle, and the six-hour-twelve-minute gap between low and high tide is half of it — a fact anyone living on that coast knows without algebra, and which the model reproduces exactly. A depth strictly between 7 and 63 would give two solutions per cycle instead, one on the way up and one on the way down.
Commit first
Answer, then rate your confidence honestly.
Predict first
Solving sine cubed x equals 9 sine x, is it legitimate to divide both sides by sine x?
Correct: No — sine x is zero at exactly the equation's solutions, so dividing discards all of them.
\[ \sin x(\sin^2x-9) = 0 \;\Longrightarrow\; \sin x = 0 \]
Why: Dividing leaves sine squared x equal to 9, which has no solutions at all, so the division has turned a solvable equation into an unsolvable one. Factoring instead keeps every case, and the zero-product property then reveals that sine x equal to 0 is where all the solutions live. Dividing by a variable expression is legal only when you know it is nonzero, and in trigonometric equations you almost never do.
Explain it
They have solved linear and quadratic equations and expect a finite answer.
Discussion prompt
In four sentences or fewer, explain why sine x equals one half has infinitely many solutions.
Hint: Think about going round the circle again.
Answer:
Sine measures a height as you walk round a circle, and there are two places on each lap where that height is one half. Then you go round again and hit both of them again.
So the solutions come in pairs, one pair per lap, forever. That is why the answer is written as a formula with an n in it rather than as a list.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the second solution, use pi minus the first for sine and 2 pi minus it for cosine. For square roots, write the plus-or-minus before doing anything else. For factoring, never divide by anything containing the variable. For extraneous solutions, substitute every candidate into the original equation whenever you have squared or multiplied through.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a trigonometric equations page. Top left: solve one basic equation, drawing the unit circle with both solutions marked and writing the general solution beneath, and note which one the inverse function would have given you. Top right: solve one equation on a stated interval, listing the general solution first and then the values of n you kept. Middle: solve one equation by factoring, showing every factor and marking clearly which ones are impossible and why. Bottom left: solve one equation in quadratic form, checking each root against the function's range before taking any inverse. Bottom right: solve one equation that requires squaring, and check all candidates in the original equation, writing out the arithmetic for the one that fails.
If any solution on your page came from dividing by a trigonometric expression, redo it by factoring: the discarded solutions are usually the ones the problem wanted.
Recap
Five things, and equations rather than identities.
| If you see | Then |
|---|---|
| An isolated function equal to a number | Inverse, then symmetry, then add periods |
| A squared function | Take both square roots |
| A common trigonometric factor | Factor; never divide |
| A quadratic in the function | Factor or use the formula, then check the range |
| A value outside -1 to 1 for sine or cosine | That case contributes nothing |
| A step that squared both sides | Check every candidate in the original |
Lesson 14.5 turns from solving to modelling: writing a trigonometric function from a maximum, a minimum and a period taken from real data.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.4 Solve Trigonometric Equations §14.4, pp. 931-935 — everything on these slides traces back here
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