Trigonometric equations and inequalities — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Solving linear trigonometric equations, quadratic trigonometric equations, and trigonometric inequalities on both restricted and general domains, using inverse trigonometric functions and unit circle symmetry to identify all solutions.

You should already know: Unit circle definitions of sine, cosine, and tangent. Pythagorean and reciprocal trigonometric identities. How to evaluate inverse trigonometric functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Trigonometric equations and inequalities?

Trigonometric equations and inequalities are algebraic statements involving one or more trigonometric functions that require finding all input values that satisfy the given relationship. Unlike polynomial or rational equations, trigonometric functions are periodic, meaning most non-restricted problems have infinitely many solutions, which are written as a general solution leveraging the known period of the function. On the AP Precalculus exam, questions almost always specify a restricted domain (most commonly $0\le x<2\pi$) but may also ask for a general solution for all real inputs, or require you to count the number of solutions in a given interval. Per the AP Precalculus Course and Exam Description (CED), this topic accounts for approximately 12% of Unit 3 (Trigonometric and Polar Functions) and appears in both multiple-choice (MCQ) and free-response (FRQ) sections. It is often combined with other topics like trigonometric identities, function modeling, and graph analysis to create multi-step exam problems.

2. Solving Linear Trigonometric Equations

A linear trigonometric equation is an equation where the trigonometric term is only raised to the first power, with the standard form: $$a\sin (bx+c)=d\text{or}a\cos (bx+c)=d\text{or}a\tan (bx+c)=d$$ where $a,b,c,d$ are real constants. The core solving strategy follows four key steps: first, isolate the trigonometric term to one side of the equation; second, check if the right-hand side is within the range of the trigonometric function (e.g., $|\sin \theta |\le 1$, so if $|d/a|>1$, there are no real solutions); third, find the reference solution for $\theta =bx+c$ using the unit circle or inverse trigonometry; fourth, use symmetry of the trigonometric function to find all solutions for $\theta$, then solve for $x$, then filter solutions to fit the required domain. For general solutions, we add a periodicity term: $2\pi n$ (for all integers $n$) for sine and cosine (which have period $2\pi$) and $\pi n$ for tangent (which has period $\pi$).

Worked Example

Find all solutions to $2\sin \left(3x-\frac{\pi }{6}\right)=1$ on the domain $0\le x<2\pi$.

1. Isolate the trigonometric term: divide both sides by 2, get $\sin \left(3x-\frac{\pi }{6}\right)=\frac{1}{2}$. The right-hand side $1/2$ is between $-1$ and $1$, so solutions exist.
2. Find solutions for $\theta =3x-\pi /6$: sine equals $1/2$ at $\theta =\frac{\pi }{6}+2\pi n$ and $\theta =\frac{5\pi }{6}+2\pi n$ for all integers $n$, by unit circle symmetry.
3. Substitute back $\theta =3x-\pi /6$ and solve for $x$: First solution set: $3x-\frac{\pi }{6}=\frac{\pi }{6}+2\pi n⟹3x=\frac{\pi }{3}+2\pi n⟹x=\frac{\pi }{9}+\frac{2\pi n}{3}$ Second solution set: $3x-\frac{\pi }{6}=\frac{5\pi }{6}+2\pi n⟹3x=\pi +2\pi n⟹x=\frac{\pi }{3}+\frac{2\pi n}{3}$
4. Filter for $0\le x<2\pi$: test integer values of $n$ to get all valid $x$: $\frac{\pi }{9},\frac{7\pi }{9},\frac{13\pi }{9},\frac{\pi }{3},\pi ,\frac{5\pi }{3}$, for 6 total solutions.

Exam tip: Always check the range of the isolated trigonometric term first—if it's outside the possible range (e.g., $\cos x=2$), you can immediately conclude no real solutions exist, saving time on MCQs.

3. Solving Quadratic Trigonometric Equations

A quadratic trigonometric equation can be rewritten as a quadratic polynomial in one trigonometric function, with the standard form: $$a{f}^2(\theta )+bf(\theta )+c=0$$ where $f(\theta )$ is $\sin \theta$, $\cos \theta$, or $\tan \theta$, and $a\ne 0$. To solve these, you first use trigonometric identities to combine like terms and get the equation into this standard form—most often, this uses the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$ to convert mixed terms (e.g., an equation with both ${\sin }^2\theta$ and $\cos \theta$) into a single-variable quadratic. Next, solve for $f(\theta )$ by factoring or the quadratic formula. Each solution for $f(\theta )$ gives a separate linear trigonometric equation, which you solve using the method for linear equations, then filter all solutions to the required domain. Always check that any solution for $f(\theta )$ is within the range of the trigonometric function, just like with linear equations.

Worked Example

Find all solutions to $2{\cos }^{2}x-3\sin x=0$ for $0\le x<2\pi$.

1. Use the Pythagorean identity to rewrite in terms of $\sin x$: $2(1-{\sin }^{2}x)-3\sin x=0$, which simplifies to $2-2{\sin }^{2}x-3\sin x=0$. Rearrange into standard quadratic form: $2{\sin }^{2}x+3\sin x-2=0$.
2. Factor the quadratic: let $u=\sin x$, so we have $2u^2+3u-2=(2u-1)(u+2)=0$. This gives solutions $u=\frac{1}{2}$ and $u=-2$.
3. Solve each linear trigonometric equation: first, $\sin x=-2$ is impossible, since the range of sine is $[-1,1]$, so discard this solution. Second, $\sin x=\frac{1}{2}$ has solutions $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$ on $0\le x<2\pi$.
4. Verify by plugging back into the original equation: for $x=\pi /6$, $2(\sqrt{3}/2{)}^2-3(1/2)=2(3/4)-3/2=3/2-3/2=0$, which checks out.

Exam tip: When you use an identity that can introduce extraneous solutions (e.g., squaring both sides of an equation), always verify your solutions by plugging them back into the original equation to eliminate any invalid solutions.

4. Solving Trigonometric Inequalities

A trigonometric inequality is a statement requiring all inputs $x$ that make a trigonometric expression greater than, less than, or equal to a given constant. The solving strategy leverages the continuity and periodicity of trigonometric functions, and follows three core steps: first, solve the corresponding equality (replace the inequality sign with an equals sign) to find all critical points (where the expression crosses the constant value) in the desired domain. Second, sort these critical points in increasing order on the interval. Third, test the sign of the trigonometric expression in each open interval between consecutive critical points (including below the first and above the last), and keep any interval where the expression satisfies the original inequality. For periodic functions, you can also use the unit circle to quickly identify the region where the inequality holds without testing every interval, which saves time.

Worked Example

Find all solutions to $\cos (2x)\ge \frac{1}{2}$ on the domain $0\le x<2\pi$.

1. First solve the corresponding equality $\cos (2x)=1/2$. Let $\theta =2x$, so $\cos \theta =1/2$, which has solutions $\theta =\frac{\pi }{3}+2\pi n$ and $\theta =\frac{5\pi }{3}+2\pi n$. Substitute back: $2x=\frac{\pi }{3}+2\pi n⟹x=\frac{\pi }{6}+\pi n$, and $2x=\frac{5\pi }{3}+2\pi n⟹x=\frac{5\pi }{6}+\pi n$.
2. Filter critical points for $0\le x<2\pi$: sorted order, we get $x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$.
3. Use the unit circle to find where $\cos \theta \ge 1/2$: cosine is greater than or equal to $1/2$ in the interval $-\frac{\pi }{3}+2\pi n\le \theta \le \frac{\pi }{3}+2\pi n$ for all integers $n$. Substitute back $\theta =2x$, so $-\frac{\pi }{3}+2\pi n\le 2x\le \frac{\pi }{3}+2\pi n⟹-\frac{\pi }{6}+\pi n\le x\le \frac{\pi }{6}+\pi n$.
4. Filter for $0\le x<2\pi$: for $n=0$, we get $0\le x\le \frac{\pi }{6}$; for $n=1$, we get $\frac{5\pi }{6}\le x\le \frac{7\pi }{6}$; for $n=2$, we get $\frac{11\pi }{6}\le x<2\pi$. Combining these, the solution is $0\le x\le \frac{\pi }{6}\cup \frac{5\pi }{6}\le x\le \frac{7\pi }{6}\cup \frac{11\pi }{6}\le x<2\pi$.

Exam tip: For sine and cosine inequalities on the unit circle, remember that cosine corresponds to the $x$-coordinate and sine corresponds to the $y$-coordinate, so you can immediately see which quadrants satisfy the inequality without testing each interval.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to divide the periodicity term by the coefficient of $x$, e.g., writing the general solution of $\sin (3x)=1/2$ as $x=\pi /6+2\pi n$ instead of $x=\pi /18+2\pi n/3$. Why: Students often only add the periodicity for the original trigonometric function and forget to adjust for the horizontal stretch. Correct move: After solving for $\theta =bx$, divide the entire solution (including the periodicity term) by $b$ to get the solution for $x$.
• Wrong move: Only finding one solution per period for sine or cosine, e.g., only writing $x=\pi /6$ for $\sin x=1/2$ on $0\le x<2\pi$, missing $x=5\pi /6$. Why: Students stop after finding the reference angle from the inverse trigonometric function, and forget sine and cosine take each value twice per period. Correct move: After finding the first solution from inverse trigonometry, always use unit circle symmetry to find the second solution in $[0,2\pi )$.
• Wrong move: Adding $2\pi n$ instead of $\pi n$ to the general solution of tangent equations. Why: Students memorize $2\pi$ periodicity from sine and cosine, and forget tangent has a shorter period of $\pi$. Correct move: Always note which trigonometric function you are working with, and add $\pi n$ for tangent, $2\pi n$ for sine and cosine.
• Wrong move: Keeping out-of-range solutions when solving quadratic trigonometric equations, e.g., keeping $u=-2$ as a valid solution for $\sin x=-2$. Why: Students focus on factoring the quadratic and forget to check if each solution is within the range of the trigonometric function. Correct move: After solving for the trigonometric function, check if any solution for sine or cosine is between $-1$ and $1$ before proceeding to find $x$.
• Wrong move: Mismatching endpoint inclusion for inequalities, e.g., including endpoints for a strict inequality $f(x)>0$. Why: Students rush to write the solution and don't check the original inequality sign. Correct move: Always match the inequality sign: include endpoints if the inequality is non-strict ($\le ,\ge$), exclude endpoints if it is strict ($<,>$).
• Wrong move: Not checking for extraneous solutions after squaring both sides of a trigonometric equation, e.g., keeping $x=3\pi /2$ as a solution to $\sin x+\cos x=1$ after squaring. Why: Squaring introduces solutions that make the two sides negatives of each other, which satisfy the squared equation but not the original. Correct move: After solving any equation where you squared both sides, plug every solution back into the original equation to discard invalid solutions.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

How many solutions does $3\tan \left(2x+\frac{\pi }{4}\right)=3\sqrt{3}$ have on the interval $0\le x<2\pi$? A) $2$ B) $3$ C) $4$ D) $5$

Worked Solution: First, isolate the tangent term by dividing both sides by 3: $\tan \left(2x+\frac{\pi }{4}\right)=\sqrt{3}$. The general solution for the argument is $2x+\frac{\pi }{4}=\frac{\pi }{3}+\pi n$ for all integers $n$, since tangent has period $\pi$. Solve for $x$: $2x=\frac{\pi }{12}+\pi n⟹x=\frac{\pi }{24}+\frac{\pi n}{2}$. Testing integer values of $n$, we get valid solutions for $n=0,1,2,3$, all of which are less than $2\pi$. For $n=4$, $x=\frac{\pi }{24}+2\pi >2\pi$, which is outside the domain. This gives 4 total solutions. Correct answer: $\boxed{C}$

Question 2 (Free Response)

Consider the function $f(x)=2{\sin }^{2}x+3\sin x-2$ for $0\le x<2\pi$. (a) Factor $f(x)$ and find all values of $x$ where $f(x)=0$. (b) Identify all intervals on $0\le x<2\pi$ where $f(x)>0$. (c) How many solutions does the equation $f(x)=2$ have on the domain $0\le x<2\pi$? Justify your answer.

Worked Solution: (a) Let $u=\sin x$, so $f(x)=2u^2+3u-2=(2u-1)(u+2)=(2\sin x-1)(\sin x+2)$. Set equal to zero: $\sin x=-2$ has no real solutions, so we only solve $\sin x=1/2$. This gives solutions $\boxed{x=\frac{\pi }{6}}$ and $\boxed{x=\frac{5\pi }{6}}$ on $0\le x<2\pi$.

(b) The critical points are $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$. Testing the sign of $f(x)$: for $0\le x<\frac{\pi }{6}$, $f(x)<0$; for $\frac{\pi }{6}<x<\frac{5\pi }{6}$, $f(x)>0$; for $\frac{5\pi }{6}<x<2\pi$, $f(x)<0$. So $f(x)>0$ on the interval $\boxed{\left(\frac{\pi }{6},\frac{5\pi }{6}\right)}$.

(c) Set $f(x)=2$: $2{\sin }^{2}x+3\sin x-2=2⟹2{\sin }^{2}x+3\sin x-4=0$. Use the quadratic formula: $\sin x=\frac{-3\pm \sqrt{9+32}}{4}=\frac{-3\pm \sqrt{41}}{4}$. $\sqrt{41}\approx 6.4$, so $\frac{-3+6.4}{4}\approx 0.85$, which is between $-1$ and $1$, giving two solutions for $x$. The other solution is $\approx -2.35$, which is outside the range of sine. Total solutions: $\boxed{2}$.

Question 3 (Application / Real-World Style)

A mass bouncing vertically on a spring oscillates around its equilibrium position. Its height above the ground (in centimeters) at time $t$ (in seconds) is modeled by $y(t)=5\cos \left(\frac{\pi }{2}t\right)+10$, for the interval $0\le t<10$ seconds. Find all times $t$ in this interval when the mass is above 12.5 cm from the ground.

Worked Solution: Set up the inequality for the height requirement: $5\cos \left(\frac{\pi }{2}t\right)+10>12.5$. Isolate the cosine term: $5\cos \left(\frac{\pi }{2}t\right)>2.5⟹\cos \left(\frac{\pi }{2}t\right)>0.5$. Solve the corresponding equality to find critical points: $\cos \left(\frac{\pi }{2}t\right)=0.5$ gives solutions $\frac{\pi }{2}t=\frac{\pi }{3}+2\pi n$ or $\frac{\pi }{2}t=\frac{5\pi }{3}+2\pi n$, so $t=\frac{2}{3}+4n$ or $t=\frac{10}{3}+4n$. The inequality $\cos \theta >0.5$ holds for $-\frac{\pi }{3}+2\pi n<\theta <\frac{\pi }{3}+2\pi n$, which translates to $-\frac{2}{3}+4n<t<\frac{2}{3}+4n$. Filtering for $0\le t<10$, the solution is $\boxed{0\le t<\frac{2}{3}\cup \frac{10}{3}<t<\frac{14}{3}\cup \frac{22}{3}<t<10}$ seconds. In context, this means the mass is above 12.5 cm from the ground during these three intervals in the first 10 seconds of oscillation.

7. Quick Reference Cheatsheet

8. What's Next

Mastering trigonometric equations and inequalities is a prerequisite for nearly all remaining topics in Unit 3 of AP Precalculus, starting with advanced modeling with periodic functions. You will use the solving techniques from this chapter to find when a periodic model reaches a specific threshold, as in the spring oscillation example here, which is a common multi-step FRQ prompt. Next, you will apply these skills when solving parametric equations involving trigonometric functions, and when finding intersection points of polar curves—this requires solving trigonometric equations to find all angles where two curves meet. Without the ability to correctly find all solutions to a trigonometric equation on a restricted domain, you will struggle to count intersection points correctly or find all valid solutions to real-world modeling problems. This topic also builds the foundation for solving calculus problems involving trigonometric functions, which relies on the same identity-based rewriting and solution strategies you learned here.

Modeling with periodic functions Parametric functions and models Polar coordinates and curves