3.8 Solving trigonometric equationsIB Maths: Analysis and Approaches SL: Revision notes

A trigonometric equation usually has many solutions, because the functions are periodic. You only need the solutions in the given interval (the general solution is not required).

Method for $\sin x = k$, $\cos x = k$ or $\tan x = k$:

1. Find the principal value (from exact values or $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ on the GDC).
2. Use symmetry to find the second value in one cycle: $\sin x = \sin(\pi - x)$, $\cos x = \cos(2\pi - x)$ (or $\cos(-x)$), $\tan x = \tan(\pi + x)$.
3. Add or subtract whole periods ($2\pi$ for sin and cos, $\pi$ for tan) until you leave the interval.

Example: $2\sin x = 1$, $0 \le x \le 2\pi$: $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.

Substitute $u$ for the inside expression and transform the interval first. For $2\sin 2x = \sqrt{3}$ with $0 \le x \le 2\pi$, let $u = 2x$, so $0 \le u \le 4\pi$. Then $\sin u = \frac{\sqrt{3}}{2}$ gives $u = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}$, and $x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$.

In general, $\sin bx = k$ has $b$ times as many solutions as $\sin x = k$ over the same interval of $x$.

If an equation contains $\sin^2 x$ and $\sin x$ (or the cosine or tangent versions), treat it as a quadratic in that function. Use $\cos^2 x + \sin^2 x = 1$ or a double angle identity so that only one function appears.

Example: $2\cos^2 x = 3\sin x$ becomes $2\sin^2 x + 3\sin x - 2 = 0$, i.e. $(2\sin x - 1)(\sin x + 2) = 0$. Then $\sin x = \frac{1}{2}$ (solutions $\frac{\pi}{6}, \frac{5\pi}{6}$), and $\sin x = -2$ is rejected because $-1 \le \sin x \le 1$.

With $\cos 2x$ present, choose the form that matches the rest: $\cos 2x = 1 - 2\sin^2 x$ if the equation also has $\sin x$.

When there is no neat analytic method, or the numbers are not exact, solve graphically: graph both sides (or the difference) on the GDC over the given interval and find each intersection or zero. State the answers to 3 significant figures.

Graphs also show how many solutions to expect, and whether an inequality such as $D(t) > 13.5$ holds between or outside the two solutions.

In context, interpret the solutions: $t = 110.4$ and $t = 232.1$ in a daylight model means days 111 to 232 inclusive have more than 13.5 hours.