Solves trigonometric equations by isolating a function, by factoring, and by using identities to reduce to a single function. Handles the infinitely many solutions periodicity produces, the extra solutions an inside coefficient creates, and the extraneous ones that squaring introduces.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 7 — Trigonometric Identities and Equations
§7.5 Solving Trigonometric Equations, pp. 844-861
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 844-861 — the pages these objectives are drawn from
Warm-up
An algebraic equation has finitely many solutions. A trigonometric one usually does not.
Discussion prompt
How many angles satisfy the equation that the sine equals one half?
Hint: Think about one turn first, then all turns.
Answer:
In one full turn there are two: one in the first quadrant and one in the second, since the sine is positive in both.
But adding a full turn to either gives another solution, and that can be repeated indefinitely — so there are infinitely many in total.
So a complete answer has two parts: the solutions within one turn, and the statement that every full turn added to each gives another. Giving only the first part is the commonest incomplete answer in this section.
Concept
Every trigonometric equation is solved in two stages: find the solutions in a single period, then add whole periods to reach the rest.
\[ \sin\theta=\tfrac{1}{2} \;\Longrightarrow\; \theta=\tfrac{\pi}{6}+2\pi k \;\text{ or }\; \tfrac{5\pi}{6}+2\pi k \]
The first stage is where the trigonometry is: an inverse gives one angle and symmetry gives the rest within the turn. The second stage is mechanical, but omitting it leaves the answer incomplete whenever the question asks for all solutions.
Figure (svg): A unit circle with two terminal sides sharing the same sine value, one in the first quadrant and one in the second
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 844-848
Section
Section 1
Concept
When one trigonometric function appears, isolate it exactly as you would isolate a variable, then convert the resulting value into angles.
Until the last two steps this is ordinary algebra with the sine standing where a variable would. Naming it as a temporary variable — letting u be the sine — makes that explicit and is a genuine help when the algebra is more than one step.
Figure (svg): A unit circle with two terminal sides sharing the same sine value, one in the first quadrant and one in the second
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 844-849
Picture it
A horizontal line at one half meets the circle twice.
Figure (svg): A unit circle with two terminal sides sharing the same sine value, one in the first quadrant and one in the second
The inverse returns only the first-quadrant angle. Reflecting across the vertical axis gives the second, and the pair then repeats every full turn.
Worked example
Ordinary algebra, then the trigonometry.
\[ \text{Solve } 2\cos\theta+1=0 \text{ for } 0\le\theta<2\pi. \]
Isolate the cosine
Why: Subtract and divide.
\[ \cos \theta = -\frac{1}{2} \]
Find the reference angle
Why: Where the cosine has size one half.
\[ \frac{\pi}{3} \]
Identify the quadrants
Why: The cosine is negative in QII and QIII.
Write both angles
Why: From the reference angle.
\[ 2 \pi / 3\text{ and } 4 \pi / 3 \]
Figure (svg): A unit circle with two terminal sides sharing the same sine value, one in the first quadrant and one in the second
\[ \theta=\tfrac{2\pi}{3},\;\tfrac{4\pi}{3} \]
Verify: substitute one solution
Why: At two pi over three the cosine is negative one half, so twice it plus one is zero — confirming that solution. Both lie in the required interval, and a cosine equation with a value strictly between negative one and one always has exactly two solutions per turn.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 845-847
Prediction
The cosine of an angle is 0.3, with the angle in one full turn.
Predict first
How many solutions are there?
Correct: Two.
Why: A horizontal line strictly between negative one and one crosses the cosine graph twice per period. The two solutions sit in the quadrants where the cosine has the required sign, symmetric about the horizontal axis.
Worked example
The interval is unbounded this time.
\[ \text{Solve } \tan\theta=1 \text{ for all } \theta. \]
Find the reference solution
Why: By the inverse tangent.
\[ \frac{\pi}{4} \]
Recall the tangent's period
Why: Half a turn, not a full one.
Note there is one solution per period
Why: Not two.
Write the general solution
Why: Add multiples of the period.
\[ \frac{\pi}{4} + \pi k \]
Figure (svg): The solution to Worked example give all solutions shown as a ladder of expressions, one row per legal move
\[ \theta=\tfrac{\pi}{4}+\pi k \]
Verify: check a second solution
Why: Five pi over four is a half turn later, and its tangent is also 1 since both coordinates are negative there. One solution per period rather than two is a feature of the tangent, whose period is already the distance between its repeated values.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 847-849
Trap
\[ \sin\theta=\tfrac{1}{2} \;\Longrightarrow\; \theta=\tfrac{\pi}{6} \]
Report the calculator's output as the solution
Why: The inverse returns one angle and it is taken to be the complete answer.
The second-quadrant solution and every full-turn repeat are omitted.
The inverse is single-valued by design, so it returns one angle out of infinitely many. Its output is the starting point, not the answer.
Reflect to find the other solution in the turn — subtract from pi for the sine — then add whole periods to both.
Read the question to see which is wanted. 'On the interval' asks for the solutions in a turn; 'all solutions' asks for the general form with the periodic term.
Faded example
The sine equals one half.
Fill in the blanks
\theta=\frac22+___\pi k \quad\text___\quad \theta=\frac______+___\pi k
Why: The sine's period is a full turn, so full turns are added to each of the two solutions in a period. Both families are needed; giving one leaves out half the solutions.
Sorting
It depends on the function.
Sort into buckets
Sort each equation.
Step zero
You face a trigonometric equation with a single function in it.
Discussion prompt
What do you do before any trigonometry?
Hint: What would you do if it were a variable?
Answer:
Isolate the function using ordinary algebra, exactly as you would isolate a variable. Nothing trigonometric is needed for this part.
Only once it stands alone equal to a number does an inverse have anything to act on. Applying one earlier has no meaning.
Naming it as a temporary variable makes this explicit — let u be the sine, solve for u, then convert u back into angles. That separation keeps the algebra and the trigonometry from getting tangled.
Section
Section 2
Concept
When two terms share a trigonometric factor, bring everything to one side and factor. Dividing by the shared factor discards every solution that makes it zero.
This is exactly the discipline from §3.6 with polynomial equations, and the reason is the same: division by a quantity that may be zero is not a legal step, and the lost solutions disappear without any warning.
Figure (svg): Four cards giving the standard strategies for solving a trigonometric equation
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 849-853
Picture it
Factoring is the second card, and the one that protects solutions.
Figure (svg): Four cards giving the standard strategies for solving a trigonometric equation
Each card matches a structural feature of the equation. Factoring applies whenever terms share a factor or the equation is quadratic in one function.
Worked example
Two terms share a sine.
\[ \text{Solve } 2\sin\theta\cos\theta=\sin\theta \text{ for } 0\le\theta<2\pi. \]
Bring everything to one side
Why: Subtract the right.
\[ 2 \sin \cos - \sin = 0 \]
Factor out the sine
Why: Common to both terms.
\[ \sin(2 \cos - 1) = 0 \]
Set each factor to zero
Why: Two equations.
\[ \sin = 0\text{ or } \cos = \frac{1}{2} \]
Solve each
Why: In the given interval.
\[ 0, \pi\text{ and } \frac{\pi}{3}, 5 \pi / 3 \]
Figure (svg): Four cards giving the standard strategies for solving a trigonometric equation
\[ \theta=0,\;\tfrac{\pi}{3},\;\pi,\;\tfrac{5\pi}{3} \]
Verify: check what division would have cost
Why: Dividing by the sine at the start would have given only the cosine equation and its two solutions, silently losing zero and pi. Substituting zero into the original gives zero equals zero, so those are genuine solutions that factoring preserved.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 850-852
Sorting
Some moves lose solutions.
Sort into buckets
Sort each step.
Worked example
Quadratic in the cosine.
\[ \text{Solve } 2\cos^2\theta-\cos\theta-1=0 \text{ for } 0\le\theta<2\pi. \]
Recognise the form
Why: A quadratic in the cosine.
\[ \text{let } u = \cos \]
Factor
Why: As an ordinary quadratic.
\[ (2 u + 1) (u - 1) = 0 \]
Solve for the cosine
Why: Two values.
\[ \cos = -\frac{1}{2}\text{ or } 1 \]
Convert to angles
Why: Two solutions and one.
\[ 2 \pi / 3, 4 \pi / 3, 0 \]
Figure (svg): The solution to Worked example factor a quadratic shown as a ladder of expressions, one row per legal move
\[ \theta=0,\;\tfrac{2\pi}{3},\;\tfrac{4\pi}{3} \]
Verify: check the count
Why: The value negative one half gives two angles and the value 1 gives only one, since the cosine reaches 1 at a single point per turn. Three solutions rather than four is correct, and noticing that an extreme value gives one angle rather than two is a useful check.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 852-853
Error analysis
A student solves an equation with a shared cosine.
Annotate
On: \( \sin\theta\cos\theta=2\cos\theta \;\Longrightarrow\; \sin\theta=2 \)
Here the division is doubly damaging: it loses the only solutions there are, since the remaining equation asks for a sine of 2 which is impossible. Factoring turns an apparently unsolvable equation into one with two solutions.
Faded example
A shared tangent.
Fill in the blanks
\tan^2\theta=\tan\theta \;\to\; \tan\theta(\tan\theta-1)=0 \;\to\; \tan\theta=0 \text1 ___
Why: Bringing the right side across and factoring gives two equations rather than one. Dividing by the tangent would have kept only the second and lost every angle where the tangent is zero.
Prediction
An equation has a common factor of the sine on both sides.
Predict first
What happens if you divide by it?
Correct: Every solution where the sine is zero is lost.
Why: Division is only legal by a nonzero quantity, and the angles making the sine zero are precisely the ones excluded. They are usually genuine solutions of the original, and nothing in the reduced equation records that they were removed.
Explain it to yourself
The rule is the same as for polynomial equations.
Discussion prompt
Explain why factoring is required rather than merely preferred.
Hint: What does division assume?
Answer:
Dividing both sides by an expression assumes that expression is not zero. For a trigonometric function that assumption is false at infinitely many angles.
Those angles are typically solutions, and after the division there is no trace that they existed. The reduced equation looks perfectly healthy.
Factoring keeps them: setting each factor to zero recovers exactly the solutions division would have removed. It is the same discipline as with polynomials, and the reason is identical — the zero-product property works forwards but division does not work backwards.
Section
Section 3
Concept
When two different functions appear, an identity from earlier in the chapter usually eliminates one, leaving something quadratic in the survivor.
Choosing which function to keep is worth a moment's thought. Keeping the one that already appears to the first power avoids introducing square roots, so an equation with a squared cosine and a plain sine should be converted to the sine.
Figure (svg): Four cards giving the standard strategies for solving a trigonometric equation
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 853-857
Picture it
The third card is this idea, and the fourth is its §7.4 variant.
Figure (svg): Four cards giving the standard strategies for solving a trigonometric equation
Which card applies is read off the equation: squared terms suggest a Pythagorean substitution, compound arguments suggest a double-angle identity, and a sum of two like terms suggests §7.4.
Worked example
Keep the function that appears to the first power.
\[ \text{Solve } 2\cos^2\theta+3\sin\theta=3 \text{ for } 0\le\theta<2\pi. \]
Choose which to keep
Why: The sine appears to the first power.
Substitute
Why: Cosine squared is one minus sine squared.
\[ 2(1 - \sin ^{2}) + 3 \sin = 3 \]
Collect
Why: A quadratic in the sine.
\[ 2 \sin ^{2} - 3 \sin + 1 = 0 \]
Factor and solve
Why: Two sine values.
\[ \sin = \frac{1}{2}\text{ or } 1 \]
Figure (svg): The solution to Worked example eliminate with a Pythagorean identity shown as a ladder of expressions, one row per legal move
\[ \theta=\tfrac{\pi}{6},\;\tfrac{\pi}{2},\;\tfrac{5\pi}{6} \]
Verify: check the solution count
Why: The value one half gives two angles and the value 1 gives one, for three in total. Substituting pi over two into the original gives zero plus three, which equals three — confirming that solution and the sine value of 1 that produced it.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 854-856
Prediction
An equation has a squared sine and a plain cosine.
Predict first
Which should you keep?
Correct: The cosine, since it appears to the first power.
Why: Replacing a squared function using the Pythagorean identity is clean, but replacing a first power introduces a square root and a sign ambiguity. Keeping the first-power function avoids both.
Worked example
A double-angle identity removes it.
\[ \text{Solve } \cos 2\theta+\cos\theta=0 \text{ for } 0\le\theta<2\pi. \]
Choose the cosine form
Why: The one written in the cosine.
\[ \cos 2 = 2 \cos ^{2} - 1 \]
Substitute
Why: Everything in the cosine now.
\[ 2 \cos ^{2} - 1 + \cos = 0 \]
Factor
Why: As a quadratic.
\[ (2 \cos - 1) (\cos + 1) = 0 \]
Solve each
Why: Two values.
\[ \cos = \frac{1}{2}\text{ or } -1 \]
Figure (svg): Four cards giving the standard strategies for solving a trigonometric equation
\[ \theta=\tfrac{\pi}{3},\;\pi,\;\tfrac{5\pi}{3} \]
Verify: check the form chosen
Why: Using the sine-only form of the double-angle identity would have left both a sine squared and a plain cosine, which does not reduce. Choosing the form matching the other term in the equation is what made the substitution productive.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 856-857
Trap
\[ 2\cos^2\theta+3\sin\theta=3 \;\to\; \text{convert the sine to a cosine} \]
Replace the sine using the Pythagorean identity
Why: The plain sine becomes a square root of one minus cosine squared.
A square root appears, and with it a sign ambiguity that was not there before.
Keep the function that already appears to the first power, and replace the squared one. Squares convert cleanly; first powers do not.
Here the sine is first-power and the cosine is squared, so replacing the cosine squared leaves a clean quadratic in the sine.
A square root introduced by an avoidable choice costs a sign decision and often an extraneous solution. Choosing the right direction takes one glance.
Matching
Each structural signal points at one tool.
Match the pairs
Why: Reading the equation's structure picks the tool. The fourth is the fallback that always applies when no more specific signal is present, and it usually reveals which of the others to use next.
Faded example
Substituting for the squared sine.
Fill in the blanks
2\sin^2\theta+\cos\theta=1 \;\to\; 2(1-\cos^2}\theta)+\cos\theta=1
Why: Replacing the squared sine leaves everything in the cosine, and collecting gives a quadratic. The plain cosine was kept because a first-power term converts badly.
Explain it
Most equations here need one function eliminated.
Discussion prompt
Explain to a classmate how to decide which one goes.
Hint: Which conversions are clean?
Answer:
Eliminate the squared one, because the Pythagorean identity converts squares cleanly — one minus the other square, with no root.
Keeping a first-power term and converting it would require a square root, bringing a sign ambiguity and a real risk of extraneous solutions.
So the rule is: keep what appears to the first power, replace what appears squared. A good explanation notes this is a one-glance decision that can save the whole second half of the problem.
Section
Section 4
Concept
An equation in a compound angle has more solutions per turn than the same equation in a plain angle, because the compound angle sweeps through more turns.
The safest order is to write the compound angle's interval explicitly before solving anything. If the original interval is a full turn and the coefficient is three, the compound angle runs over three full turns, and writing that down makes the extra solutions appear naturally.
Figure (svg): A diagram showing that solving for a doubled angle produces twice as many solutions in one turn
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 857-860
Picture it
The compound angle sweeps further than the plain one.
Figure (svg): A diagram showing that solving for a doubled angle produces twice as many solutions in one turn
The count is multiplied by exactly the coefficient. Widening the interval first turns finding them into routine work rather than a hunt.
Worked example
Widen the interval before solving.
\[ \text{Solve } \sin 2\theta=\tfrac{\sqrt{3}}{2} \text{ for } 0\le\theta<2\pi. \]
Widen the interval
Why: Twice theta runs over two turns.
\[ 0 \le 2 \theta < 4 \pi \]
Solve within one turn
Why: Two solutions.
\[ \frac{\pi}{3}\text{ and } 2 \pi / 3 \]
Add a full turn to each
Why: For the second turn.
\[ 7 \pi / 3\text{ and } 8 \pi / 3 \]
Divide all four by two
Why: Recover theta.
\[ \frac{\pi}{6}, \frac{\pi}{3}, 7 \pi / 6, 4 \pi / 3 \]
Figure (svg): A diagram showing that solving for a doubled angle produces twice as many solutions in one turn
\[ \theta=\tfrac{\pi}{6},\;\tfrac{\pi}{3},\;\tfrac{7\pi}{6},\;\tfrac{4\pi}{3} \]
Verify: check the count against the coefficient
Why: A plain sine equation has two solutions per turn, and the coefficient two doubles that to four — matching what was found. Substituting pi over six gives the sine of pi over three, which is root three over two, confirming one of them.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 857-859
Prediction
An equation in the sine of four theta, over one full turn.
Predict first
How many solutions?
Correct: Eight.
Why: A plain sine equation with a value strictly between negative one and one has two solutions per turn, and the coefficient four multiplies that. The compound angle sweeps four turns, each contributing two.
Worked example
The count triples.
\[ \text{Solve } \cos 3\theta=0 \text{ for } 0\le\theta<2\pi. \]
Widen the interval
Why: Three theta runs over three turns.
\[ 0 \le 3 \theta < 6 \pi \]
Solve the base equation
Why: The cosine vanishes every half turn.
\[ \frac{\pi}{2} + \pi k \]
List those below six pi
Why: Six values.
\[ \frac{\pi}{2}\text{ through } 11 \pi / 2 \]
Divide each by three
Why: Recover theta.
Figure (svg): The solution to Worked example a coefficient of three shown as a ladder of expressions, one row per legal move
\[ \theta=\tfrac{\pi}{6}+\tfrac{\pi}{3}k, \; k=0\ldots5 \]
Verify: check the spacing
Why: The solutions are pi over three apart, which is the original half-turn spacing divided by the coefficient three. Both the count tripling and the spacing shrinking by a factor of three are consequences of the same compression, so they confirm each other.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 859-860
Error analysis
A student solves an equation in a doubled angle.
Annotate
On: \( \sin 2\theta=\tfrac{1}{2} \text{ on } [0,2\pi): \; 2\theta=\tfrac{\pi}{6},\tfrac{5\pi}{6} \;\Rightarrow\; \theta=\tfrac{\pi}{12},\tfrac{5\pi}{12} \)
Half the solutions were lost. Writing the compound angle's widened interval down before solving is what prevents this, and it takes one line.
Faded example
Solving for the cosine of five theta over one turn.
Fill in the blanks
0\le\theta<2\pi \;\Longrightarrow\; 0\le 5\theta<10\pi, \text5 ___ \text___
Why: Multiplying the interval's endpoints by the coefficient gives the range the compound angle actually covers. Writing this first is what makes the extra solutions appear rather than having to be noticed afterwards.
Sorting
Only an inside coefficient multiplies the count.
Sort into buckets
Sort each equation, over one full turn.
Explain it to yourself
An inside coefficient multiplies the number of solutions.
Discussion prompt
Explain why, and why an outside one does not.
Hint: What does each coefficient act on?
Answer:
An inside coefficient acts on the angle before the function does, so as theta crosses one turn the compound angle crosses several. Each of those turns contributes its own solutions.
An outside coefficient acts on the output after the function has been evaluated. It changes what value is being matched but not how far the angle travels.
So only the inside one multiplies the count, and it multiplies it by exactly itself. Writing the widened interval first turns this from something to remember into something already on the page.
Section
Section 5
Concept
Squaring both sides can create solutions that do not satisfy the original, and multiplying by a denominator can create ones outside the domain. Both require every candidate to be tested.
Extraneous solutions arise because squaring destroys sign information: an equation saying two quantities are negatives of each other becomes one saying they are equal in size, which is a weaker statement admitting more angles.
Figure (svg): A contrast between operations that preserve the solution set and operations that require checking afterwards
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 855-861
Picture it
The left column changes the solution set.
Figure (svg): A contrast between operations that preserve the solution set and operations that require checking afterwards
The right column is safe because every step is reversible. The left column is not, which is why a check is part of the method rather than an optional courtesy.
Worked example
Squaring creates a candidate that fails.
\[ \text{Solve } \sin\theta+1=\cos\theta \text{ for } 0\le\theta<2\pi. \]
Square both sides
Why: To remove the mixed functions.
\[ \sin ^{2} + 2 \sin + 1 = \cos ^{2} \]
Substitute for the cosine squared
Why: By the Pythagorean identity.
\[ \sin ^{2} + 2 \sin + 1 = 1 - \sin ^{2} \]
Collect and factor
Why: A quadratic in the sine.
\[ 2 \sin(\sin + 1) = 0 \]
Solve for candidates
Why: Three angles.
\[ 0, \pi, 3 \pi / 2 \]
Test each in the original
Why: One fails.
Figure (svg): A contrast between operations that preserve the solution set and operations that require checking afterwards
\[ \theta=0,\;\tfrac{3\pi}{2} \]
Verify: check the rejected candidate
Why: At pi the left side is zero plus one, which is 1, and the right side is negative 1. They are negatives of each other, which is exactly what squaring cannot distinguish — so pi satisfies the squared equation but not the original. Testing every candidate is what catches it.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 856-858
Prediction
Squaring both sides of an equation.
Predict first
What information is lost?
Correct: The sign, so negatives become indistinguishable from positives.
Why: After squaring, an equation stating that two quantities are equal cannot be distinguished from one stating they are negatives. That weaker statement admits angles the original excluded, which is exactly what an extraneous solution is.
Worked example
A candidate can lie outside the original's domain.
\[ \text{Solve } \tan\theta=\sin\theta \text{ for } 0\le\theta<2\pi. \]
Rewrite the tangent
Why: As sine over cosine.
\[ \sin / \cos = \sin \]
Bring to one side and factor
Why: Multiply through and collect.
\[ \sin(1 - \cos) = 0 \]
Solve each factor
Why: Three candidates.
\[ 0, \pi\text{ and } 0 \]
Check the domain
Why: The tangent must be defined.
Figure (svg): The solution to Worked example a domain restriction shown as a ladder of expressions, one row per legal move
\[ \theta=0,\;\pi \]
Verify: check the excluded angles
Why: The tangent is undefined at pi over two and three pi over two, so any candidate there would have to be rejected regardless of the algebra. Neither solution falls there, so both survive — but the check is still needed, since multiplying through by the cosine is the step that could have admitted them.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 858-861
Trap
\[ \text{test each candidate in the squared equation} \]
Substitute into the equation after squaring
Why: The candidates are verified against the most recent form.
Every candidate passes, including the extraneous ones.
Check against the original equation, before any squaring or multiplying. That is the statement the answers must satisfy.
Extraneous solutions satisfy the squared equation by construction — that is exactly why they appeared — so testing there catches nothing.
Write the original down separately before starting, so it is available at the end. Working from the bottom of the page is how candidates get checked against the wrong thing.
Sorting
Only irreversible steps do.
Sort into buckets
Sort each operation.
Faded example
Testing pi in the equation sine plus one equals cosine.
Fill in the blanks
\text1=0+1=-1, \quad \text___=\cos\pi=___, \text___
Why: The two sides are negatives of each other, which squaring cannot distinguish. The candidate satisfies the squared equation and fails the original, so it must be discarded — and only a test against the original reveals that.
Explain it
Squaring can produce answers that do not work.
Discussion prompt
Explain to a classmate where they come from and how to catch them.
Hint: What does squaring do to a minus sign?
Answer:
Squaring turns a statement that two things are equal into a statement that they are equal in size. That is weaker, so more angles satisfy it.
Any angle where the two sides are negatives of each other now qualifies, even though it failed the original. That is precisely an extraneous solution.
The only cure is to test every candidate in the original equation. A good explanation stresses that this is part of the method rather than a precaution — once you square, the check is what makes the answer correct.
Comparison
Fill the blanks from memory. The equation's shape picks the method.
Comparison matrix
| isolate | factor | reduce to one function | |
|---|---|---|---|
| when the equation has | a single function | a shared factor or a quadratic | two different functions |
| the key move | ordinary algebra, then an inverse | bring to zero and factor | substitute an identity |
| what can go wrong | giving only one solution | dividing instead of factoring | eliminating the first-power function |
| needs a final check | only the interval | only the interval | yes, if squaring was used |
The third row lists the characteristic error of each method. Knowing which one you are exposed to is more useful than a general instruction to be careful.
Pattern
Five steps, and the last is not optional.
Step 4's widening has to happen before step 3's solving, not after, or the extra solutions never appear as candidates at all.
OpenStax Algebra and Trigonometry 2e, §9.5 Solving Trigonometric Equations §9.5
Check
Factoring versus dividing.
Check your understanding
An equation has a sine on both sides as a factor. What should you do?
Answer: A
Why: Dividing by the sine discards every angle where the sine is zero, and those are usually solutions. Factoring keeps them, since setting each factor to zero recovers exactly what division would have removed.
Check
Multiple angles.
Check your understanding
Over one full turn, how many solutions does an equation in the sine of three theta typically have?
Answer: A
Why: The compound angle sweeps three full turns while theta sweeps one, and each turn contributes the usual two solutions. The coefficient multiplies the count by exactly itself.
Check
Extraneous solutions.
Check your understanding
After squaring both sides, where must each candidate be checked?
Answer: A
Why: Extraneous solutions satisfy the squared equation by construction, so testing there catches nothing. Only the original distinguishes the genuine solutions from those that arose when sign information was destroyed.
Real world
Any question about when a periodic process reaches a given level is one of these equations.
Discussion prompt
A tide is modelled by a sinusoid. Finding when the water reaches a given depth is a trigonometric equation — what does its structure tell you?
Hint: How many times a day does the tide pass a given level?
Answer:
Each tidal cycle crosses a given level twice, once rising and once falling. That is the two solutions per period, appearing as a physical fact.
And the crossings repeat every cycle, which is the periodic extension. The general solution's two families are the rising crossings and the falling ones.
So the mathematical structure is not an artefact — the two families correspond to two physically different events, and a harbour master needs the rising ones for entry and the falling ones for departure. Losing half the solutions here means missing half the safe windows.
Commit first
State your confidence along with your answer.
Predict first
Why must you factor rather than divide by a trigonometric expression?
Correct: Because it can be zero, and dividing discards those solutions.
Why: Division is legal only by a nonzero quantity, and the angles making a trigonometric expression zero are typically solutions of the equation. They vanish with no trace, which is what makes the error silent and therefore dangerous.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate solved an equation in the sine of two theta and found only two solutions. Explain what went wrong.
Hint: How far does the doubled angle travel?
Answer:
As theta sweeps one full turn, twice theta sweeps two. So the doubled angle passes each value twice as often.
They solved over one turn for the doubled angle and stopped, missing everything in the second turn. Four solutions exist, not two.
The fix is mechanical: widen the interval before solving. If theta runs to two pi then twice theta runs to four pi, and writing that down first makes the extra solutions appear as ordinary candidates rather than something to remember to look for.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth combines the two errors that cost the most marks, since both lose or add solutions silently. The second is the discipline that carries over from chapter 3 unchanged.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the four solving strategies with the structural signal that triggers each. Beside them, work one equation with a doubled angle from start to finish, showing the widened interval explicitly. In a box at the bottom, list the two operations that require a check afterwards and say what each one does to the solution set.
If your widened interval appears before any solving and your box distinguishes losing solutions from gaining them, the section's two costly errors are both accounted for.
Recap
Five things, and the last two are where the marks are lost.
| if you remember one thing | it should be this |
|---|---|
| about completeness | two solutions per turn, then add whole periods |
| about factoring | dividing by a function loses solutions silently |
| about multiple angles | widen the interval first, divide the answers last |
| about squaring | it can only add solutions, so check every candidate |
Section 7.6 closes the chapter by building sinusoidal models from data — damped and undamped — where solving equations like these answers the questions the model was built to address.
OpenStax, Precalculus, §7.5 Solving Trigonometric Equations §7.5, pp. 844-861 — everything on these slides traces back here
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