3.10 Trigonometric Equations and Inequalities

Trigonometric equations and inequalities ask you to solve for the angles that make a trig statement true. You use inverse trig functions to find a starting solution, then adjust for domain restrictions and add full periods to capture every answer.

Why This Matters for the AP Precalculus Exam

Trigonometric functions make up a large part of Unit 3, the heaviest unit on the AP Precalculus exam. Solving trig equations and inequalities pulls together the unit circle, inverse trig functions, and periodicity, so it tests several skills at once.

On the exam you may be asked to solve a trig equation by hand, find all solutions in a given interval, or solve an equation that comes from a real situation where the context limits which answers make sense. Some questions allow a graphing calculator, which you can use to find numerical solutions or intersection points, while others expect exact work with the unit circle. Clear setup and correct interval notation make your reasoning easy to follow.

Key Takeaways

• Use inverse trig functions (arcsin, arccos, arctan) to find a first solution, then check it against the function's range.
• Sine and cosine repeat every 2π (360 degrees), and tangent repeats every π (180 degrees), so write general solutions by adding full periods.
• Each inverse function has a fixed output range: arcsin gives [-π/2, π/2], arccos gives [0, π], and arctan gives (-π/2, π/2).
• The inverse function returns only one angle, so use the unit circle or symmetry to find the second solution in a full revolution.
• For inequalities, find the boundary angles first, then decide which arc of the unit circle satisfies the inequality.
• A real-world context often sets a restricted domain, which cuts the infinite list of solutions down to a few valid ones.

Solving Trigonometric Equations

Inverse trig functions undo regular trig functions, so they are your main tool for solving. If you have an equation like sin(x) = 0.5, take the arcsine of both sides to get a starting angle.

This works just like solving 3x = 6 by dividing both sides by 3. Arcsine is the inverse of sine, so applying it isolates x. For sin(x) = 0.5, arcsin(0.5) = 30 degrees, because the sine of 30 degrees is 0.5.

The catch is that the inverse function only gives back one angle. Sine is positive in both the first and second quadrants, so sin(x) = 0.5 also has a solution at 180 - 30 = 150 degrees within one revolution. You have to use the unit circle or symmetry to find that second angle yourself.

Periodicity Means Infinite Solutions

Sine, cosine, and tangent are periodic, so trig equations usually have infinitely many solutions. The equation sin(x) = 0.5 has a solution at 30 degrees, but also at 390 degrees, 750 degrees, and so on, because sine repeats every 360 degrees (2π radians).

To capture every solution, write a general form using an integer k:

• For sin(x) = 0.5, the two base angles are 30 degrees and 150 degrees, so x = 30 + 360k or x = 150 + 360k.
• Tangent repeats every 180 degrees, so a tangent equation needs only one base angle plus 180k.

This periodic repeating is also why the inverse functions have restricted ranges. Each inverse returns just one output so that it stays a true function.

Worked Example: cos(x) = -0.2

Solve cos(x) = -0.2 for x on the interval 0 to 360 degrees.

1. The cosine term is already isolated, so take the arccosine of both sides: x = arccos(-0.2).
2. A calculator gives arccos(-0.2) ≈ 101.5 degrees. This is the value inside the arccosine range [0, 180 degrees], so it is a valid solution.
3. Cosine is also negative in the third quadrant. The second solution is 360 - 101.5 = 258.5 degrees.

So on 0 to 360 degrees, x ≈ 101.5 degrees or x ≈ 258.5 degrees. The arccosine function handed you only the first angle, so you used the unit circle to find the second.

Solving Trigonometric Inequalities

Inequalities work much like equations, but instead of single angles you describe a range of angles. Start by solving the matching equation to find the boundary angles, then decide which part of the unit circle makes the inequality true.

For sin(x) > 0.5 on 0 to 360 degrees, the boundaries come from sin(x) = 0.5, which are 30 degrees and 150 degrees. Sine is above 0.5 between those two angles, so the solution is 30 degrees < x < 150 degrees.

You can use the unit circle or a graph of y = sin(x) to see which arc satisfies the inequality. Watch your calculator mode: degree mode and radian mode give different numbers, so match the mode to the problem.

Domain and Range Restrictions on Trigonometric Functions

When you use an inverse trig function, the answer it returns is locked to that function's range. For example, arcsin only outputs angles between -90 and 90 degrees, so a calculator answer that should land elsewhere needs to be adjusted using the unit circle.

The chart below lists the domain and range of sine, cosine, and tangent, along with their inverses. Square brackets mean the endpoint is included, and parentheses mean it is excluded.

How to Use This on the AP Precalculus Exam

Problem Solving

• Take the inverse trig function to get one base angle, then use the unit circle to find the second angle in a full revolution.
• Decide whether the problem wants all solutions or only those in a given interval. If it wants all of them, write a general solution with +360k (or +2πk) for sine and cosine, and +180k (or +πk) for tangent.
• Match your calculator mode to the units in the problem. Mixing degrees and radians is a common source of wrong answers.

Free Response

• Show the inverse step and the unit-circle reasoning that gives the second solution. Clear steps make your work easy to follow.
• State the interval you are solving over, and use correct interval or inequality notation in your final answer.

Common Trap

• The inverse function gives only one angle. Forgetting the second quadrant solution (for sine and cosine) is one of the most common mistakes.

Common Misconceptions

• "The inverse function gives all the solutions." It gives only one angle inside its restricted range. You still need the unit circle or periodicity to find the rest.
• "Every solution is valid." If you used an inverse function, a candidate angle must fall in that inverse's range, or you have to adjust it. In a real-world problem, the context may rule out solutions too.
• "arcsin(0.6) is 150 degrees in the second quadrant." The second-quadrant partner of arcsin(0.6) is about 180 - 36.9 = 143.1 degrees, not 150 degrees. Only use 150 degrees when the first-quadrant angle is exactly 30 degrees.
• "Sine and tangent repeat the same way." Sine and cosine repeat every 360 degrees, but tangent repeats every 180 degrees, so their general solutions use different periods.
• "Inequalities have a single answer." A trig inequality describes a range of angles, so your answer is usually an interval, not one value.

Practice Problems

1. Solve sin(x) = 0.6 on the interval 0 to 360 degrees. Round to the nearest tenth of a degree.

First-quadrant angle: arcsin(0.6) ≈ 36.9 degrees. Second solution (sine positive in quadrant II): 180 - 36.9 = 143.1 degrees.

Answer: x ≈ 36.9 degrees or x ≈ 143.1 degrees.

2. Solve the inequality cos(x) > 0.5 on the interval 0 to 360 degrees.

Boundary angles from cos(x) = 0.5: x = 60 degrees and x = 300 degrees. Cosine is greater than 0.5 near 0 degrees, so the solution is the arc from 0 to 60 degrees and from 300 to 360 degrees.

Answer: 0 degrees ≤ x < 60 degrees or 300 degrees < x ≤ 360 degrees (equivalently, the angles within 60 degrees of 0).

3. Solve tan(x) = -2 on the interval -90 to 90 degrees.

Arctangent returns values in (-90, 90 degrees), so x = arctan(-2) ≈ -63.4 degrees, which is already inside the interval.

Answer: x ≈ -63.4 degrees.

Related AP Precalculus Guides

• 3.3 Sine and Cosine Function Values
• 3.6 Sinusoidal Function Transformations
• 3.4 Sine and Cosine Function Graphs
• 3.7 Sinusoidal Function Context and Data Modeling
• 3.9 Inverse Trigonometric Functions
• 3.8 The Tangent Function