Solving trig equations

Periodic graphs cross a level many times: Because sin, cos and tan repeat, an equation like sin x = 0.5 has many solutions. A question fixes a domain (e.g. 0° ≤ x ≤ 360°) and wants all solutions in it — usually two for sin/cos.

Read the interval first: Underline the domain. The number of answers depends on it — over one full period sin = k (|k|<1) gives two solutions.

First solution, then its partner: Take the inverse for the first angle, then use symmetry for the others: sin x = k → x and 180° − x; cos x = k → x and 360° − x; tan x = k → x and x + 180°.

Negative k changes the quadrants: For cos x = −0.5 the solutions are in Q2 and Q3; sketch the unit circle (or use CAST) to place them correctly.

Stretch the interval, solve, then divide: For sin(2x) = k on 0° ≤ x ≤ 360°: let the inside be the new variable, solve over the stretched interval 0° ≤ 2x ≤ 720°, find all those solutions, then divide each by 2. A multiple angle gives more solutions.

Don't divide too early: Find all the 2x solutions across the doubled interval first — if you halve before finding them all, you lose answers.

Substitute, factor, then solve each: An equation like 2sin²x − sinx − 1 = 0 is a quadratic in sin x. Let s = sin x, factor/solve for s, then solve sin x = each value over the interval (rejecting any s outside [−1, 1]).

Use a Pythagorean swap if mixed: If the equation mixes sin² and cos (e.g. 2cos²x + sinx = …), use cos²x = 1 − sin²x first to get a single ratio, then it's a quadratic.

Graph and use intersect: On Paper 2, solve a trig equation by graphing each side and using intersect (or graphing the difference and finding zeros) over the given interval — ideal inside a sinusoidal model (e.g. 'when is the height 5 m?').