Trigonometric Equations Worksheet (with solutions) | Teaching ... - Free Printable

(a) $\cos x - \sin x = 0$

$\cos x = \sin x$

$\frac{\sin x}{\cos x} = 1$

$\tan x = 1$

$x = \frac{\pi}{4} + n\pi$, where $n$ is an integer.

(b) $2\cos^2 x = 1$

$\cos^2 x = \frac{1}{2}$

$\cos x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$

$x = \frac{\pi}{4} + n\frac{\pi}{2}$, where $n$ is an integer.

(c) $4\sin^2 x + 2 = 3$

$4\sin^2 x = 1$

$\sin^2 x = \frac{1}{4}$

$\sin x = \pm \frac{1}{2}$

$x = \frac{\pi}{6} + 2n\pi$, $x = \frac{5\pi}{6} + 2n\pi$, $x = \frac{7\pi}{6} + 2n\pi$, $x = \frac{11\pi}{6} + 2n\pi$, where $n$ is an integer.

(d) $\sin^2 x - 3\sin x + 2 = 0$

Let $u = \sin x$, then $u^2 - 3u + 2 = 0$

$(u - 1)(u - 2) = 0$

$u = 1$ or $u = 2$

$\sin x = 1$ or $\sin x = 2$

$\sin x = 2$ has no solution since $|\sin x| \leq 1$

$\sin x = 1 \Rightarrow x = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

(e) $\sin x - 2\sin x \cos x = 0$

$\sin x(1 - 2\cos x) = 0$

$\sin x = 0$ or $1 - 2\cos x = 0$

$\sin x = 0 \Rightarrow x = n\pi$, where $n$ is an integer.

$1 - 2\cos x = 0 \Rightarrow \cos x = \frac{1}{2} \Rightarrow x = \frac{\pi}{3} + 2n\pi$ or $x = \frac{5\pi}{3} + 2n\pi$, where $n$ is an integer.

So, $x = n\pi$, $x = \frac{\pi}{3} + 2n\pi$, $x = \frac{5\pi}{3} + 2n\pi$, where $n$ is an integer.