Trigonometric Equations and Inequalities

A trigonometric equation contains a trigonometric function of an unknown angle, such as 2 sin x = 1, and it usually has infinitely many solutions that repeat with the period of the function. In brief: sin x = 1/2 has the solutions x = π/6 + 2kπ and x = 5π/6 + 2kπ for every integer k. In the interval [0, 2π) there are exactly two: π/6 and 5π/6. Trigonometric inequalities are solved by reading the graph or the unit circle.

The strategy is to reduce the problem to a basic equation sin u = a, cos u = a or tan u = a, solve it using the unit circle and then add the periodic part and solve for x. This guide describes the three basic equations, then equations that need factoring, identities or substitutions, followed by inequalities, with a table of general solutions.

How do you solve the basic equations?

1. Isolate the trigonometric function: for example, sin x = a.
2. If |a| > 1 for sine or cosine, there is no solution.
3. Find the reference angle α = sin⁻¹|a|.
4. Use the signs in the quadrants to find all angles in one period.
5. Add the period to get the general solution.

Example 1: sin x = 1/2

Sine is positive in quadrants I and II with reference angle π/6. So x = π/6 and x = 5π/6 in [0, 2π). General solution: x = π/6 + 2kπ or x = 5π/6 + 2kπ.

Example 2: cos x = −√2/2

Cosine is negative in quadrants II and III with reference angle π/4: x = 3π/4 and 5π/4. General: x = ±3π/4 + 2kπ.

Example 3: tan x = √3

Tangent has period π, with the solution π/3 in [0, π). General: x = π/3 + kπ.

Example 4: no solution

sin x = 1.5 has no solution, since −1 ≤ sin x ≤ 1.

What are the general solutions?

For an integer k and the principal value α = sin⁻¹ a, cos⁻¹ a, tan⁻¹ a:

• sin x = a: x = α + 2kπ or x = π − α + 2kπ.
• cos x = a: x = ±α + 2kπ.
• tan x = a: x = α + kπ.

For an argument other than x, such as sin 2x = a, solve for 2x and then divide: 2x = α + 2kπ, so x = α/2 + kπ. The number of solutions in an interval depends on the period: sin 2x = 1/2 has four solutions in [0, 2π), since 2x ranges over [0, 4π].

Example 5: sin 2x = 1/2

2x = π/6 + 2kπ or 5π/6 + 2kπ. So x = π/12 + kπ or 5π/12 + kπ. In [0, 2π): π/12, 5π/12, 13π/12 and 17π/12.

How do you solve harder equations?

Factoring

2 sin x cos x − cos x = 0. Factor: cos x (2 sin x − 1) = 0. So cos x = 0, giving x = π/2 + kπ, or sin x = 1/2, giving x = π/6 + 2kπ and 5π/6 + 2kπ. Do not divide by cos x, because you would lose the solutions where cos x = 0.

Quadratic in one function

2 cos²x + cos x − 1 = 0. Let u = cos x: 2u² + u − 1 = 0, so (2u − 1)(u + 1) = 0, giving u = 1/2 or u = −1. Then x = ±π/3 + 2kπ or x = π + 2kπ.

Using the Pythagorean identity

2 sin²x + 3 cos x = 3. Replace sin²x by 1 − cos²x: 2 − 2 cos²x + 3 cos x − 3 = 0, so 2 cos²x − 3 cos x + 1 = 0, giving (2 cos x − 1)(cos x − 1) = 0. So cos x = 1/2 or cos x = 1: x = ±π/3 + 2kπ or x = 2kπ.

Combining sine and cosine

sin x + cos x = 1. Square both sides: 1 + 2 sin x cos x = 1, so sin 2x = 0. Then x = kπ/2. Check each in the original: x = 0 gives 0 + 1 = 1 ✓; x = π/2 gives 1 + 0 = 1 ✓; x = π gives −1 ✗; x = 3π/2 gives −1 ✗. So x = 2kπ or x = π/2 + 2kπ. Squaring can add false solutions, so always check.

Double angle

sin 2x = sin x. Using sin 2x = 2 sin x cos x: sin x (2 cos x − 1) = 0, so sin x = 0 (x = kπ) or cos x = 1/2 (x = ±π/3 + 2kπ).

How do you solve trigonometric inequalities?

Solve the related equation first to find the boundary points, then determine from the graph or the unit circle in which intervals the inequality holds.

Example 6: sin x > 1/2 for x in [0, 2π)

Boundaries: π/6 and 5π/6. The sine is above 1/2 between them, so the solution is (π/6, 5π/6). The general solution is π/6 + 2kπ < x < 5π/6 + 2kπ.

Example 7: cos x ≤ −1/2 for x in [0, 2π)

Boundaries: 2π/3 and 4π/3. On the unit circle, the x-coordinate is at most −1/2 on the arc between them. The solution is [2π/3, 4π/3].

Example 8: tan x > 1 for x in (−π/2, π/2)

Boundary: π/4. Tangent is increasing, so the solution is (π/4, π/2).

Table of basic solutions

Where do these methods come from?

Trigonometric equations arose in astronomy, when the observed angle was known and the time or the position had to be found. Medieval Islamic astronomers developed tables and interpolation methods. The use of general solutions with an integer parameter k is a modern algebraic convention that emerged with the functional treatment of trigonometry by Euler in the eighteenth century.

Where are trigonometric equations used today?

The time of day when the Sun is at a given height, the moments when the tide is above a threshold and the times at which an oscillating current passes through zero are all solutions of trigonometric equations. For a Ferris wheel with h(t) = 22 − 20 cos(πt/4), the rider is above 32 m when cos(πt/4) < −1/2, so 2π/3 < πt/4 < 4π/3 and 8/3 < t < 16/3 minutes.

Which mistakes are common?

Giving only one solution

Sine and cosine take each value (between −1 and 1) twice per period. Find both.

Forgetting the periodic part

If the problem asks for all solutions, add 2kπ (or kπ for tangent).

Dividing by a trigonometric factor

Divide only if you are sure it is not zero. Factoring is safer.

Not checking after squaring

Squaring both sides can create false solutions. Substitute the answers into the original equation.