Solving Trigonometric Equations on an Interval

You’ll be able to

Find every solution of a trigonometric equation within a specified interval
Use reference angles and quadrant signs to generate all solutions from one
Handle equations where the argument is a multiple of the variable

One solution from the inverse, the rest from symmetry

The inverse function gives one angle. Every other solution comes from two facts: within a single revolution there is usually a second angle with the same value, found by reflection; and all solutions repeat with the function's period. So the procedure is: find the reference angle, decide which quadrants have the right sign, produce both angles in one revolution, then add whole periods until you leave the interval.

Generating all solutions

sin x = c → x = α and π − α, plus 2πn · cos x = c → x = α and −α (or 2π − α), plus 2πn · tan x = c → x = α, plus πn

α is the value the inverse function returns. Tangent needs only one base solution per period because its period is π, not 2π.

Worked example

Solve 2 sin x + 1 = 0 for 0 ≤ x < 2π.

1.Isolate: sin x = −1/2.
2.Reference angle: arcsin(1/2) = π/6.
3.Sine is negative in quadrants III and IV.
4.Quadrant III: x = π + π/6 = 7π/6.
5.Quadrant IV: x = 2π − π/6 = 11π/6.
6.Both lie in [0, 2π), and no more solutions fit in one revolution.

Answer: x = 7π/6 and x = 11π/6. Taking the reference angle from the absolute value and then placing it by quadrant is more reliable than working directly with arcsin(−1/2), which returns −π/6 — a correct angle, but outside the requested interval.

Watch out

A trigonometric equation on an interval almost never has exactly one solution. Finding one and stopping is the most common way to lose points on these problems — count how many periods fit in the interval and expect roughly that many times two.

Checkpoint

How many solutions does cos x = 0.4 have on [0, 4π)?

When the argument is bx

For sin(3x) = 0.5 on [0, 2π), do not solve for x too early. Substitute u = 3x: as x runs over [0, 2π), u runs over [0, 6π) — three times as far. Solve for u throughout that longer interval, then divide every solution by 3. The compression means three times as many solutions, and forgetting to extend the u-interval is what loses two thirds of them.

Worked example

Solve sin(2x) = √3/2 for 0 ≤ x < 2π.

1.Let u = 2x. As x runs over [0, 2π), u runs over [0, 4π).
2.Reference angle: arcsin(√3/2) = π/3. Sine is positive in quadrants I and II.
3.In [0, 2π): u = π/3 and u = 2π/3.
4.Add 2π for the second revolution: u = π/3 + 2π = 7π/3 and u = 2π/3 + 2π = 8π/3. Both are below 4π. ✓
5.Divide each by 2: x = π/6, π/3, 7π/6, 4π/3.

Answer: x = π/6, π/3, 7π/6 and 4π/3 — four solutions. Solving in u over the stretched interval and dividing at the end is what produces all four; solving for x directly typically yields only the first two.

Quadratic in a trigonometric function

An equation like 2 cos²x − cos x − 1 = 0 is quadratic in cos x. Substitute u = cos x to get 2u² − u − 1 = 0, factor to (2u + 1)(u − 1) = 0, and read off u = −1/2 or u = 1. Then solve each separately. Discard any u outside [−1, 1] immediately — sine and cosine cannot produce it, so it contributes no solutions at all.

Worked example

Solve 2 cos²x − cos x − 1 = 0 for 0 ≤ x < 2π.

1.Substitute u = cos x: 2u² − u − 1 = 0.
2.Factor: (2u + 1)(u − 1) = 0, so u = −1/2 or u = 1.
3.For cos x = −1/2: reference angle π/3, cosine negative in quadrants II and III, giving x = 2π/3 and 4π/3.
4.For cos x = 1: this happens only at x = 0 in the interval.
5.Collect: x = 0, 2π/3, 4π/3.

Answer: x = 0, 2π/3 and 4π/3. Notice that cos x = 1 contributes just one solution rather than two, because the value 1 is attained at a maximum rather than at two separate crossings — endpoint values of the range always behave this way.

Checkpoint

In solving 2 sin²x + 3 sin x + 1 = 0 you find sin x = −1/2 or sin x = −1. Which contributes fewer solutions on [0, 2π)?

Answer the 2 checkpoints as you read.