Trigonometric Equations practice questions — A-Level Maths

$R = \sqrt{1 + 3} = 2$, so $f(x) = 2\sin(x + \pi/3)$. Max value: $2$.

$2\cos^2 x + \cos x - 1 = 0 \Rightarrow (2\cos x - 1)(\cos x + 1) = 0$. $\cos x = 1/2$: 2 solutions ($\pi/3, 5\pi/3$). $\cos x = -1$: 1 solution ($\pi$). Total: $3$.

$2\sin x \cos x - \sin x = 0 \Rightarrow \sin x (2\cos x - 1) = 0$. $\sin x = 0$: $x \in \{0, \pi\}$. $\cos x = 1/2$: $x \in \{\pi/3, 5\pi/3\}$. Total: $4$.

$(2\sin x - 1)(\sin x - 1) = 0$. From $\sin x = \tfrac{1}{2}$: $x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$. From $\sin x = 1$: $x = \tfrac{\pi}{2}$. Total: $3$.

$\cos u = -\tfrac{1}{2}$ on $[0, 4\pi)$ at $u = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}, \tfrac{8\pi}{3}, \tfrac{10\pi}{3}$. Dividing by $2$: $x = \tfrac{\pi}{3}, \tfrac{2\pi}{3}, \tfrac{4\pi}{3}, \tfrac{5\pi}{3}$. Four solutions.

$\arctan(1) = \pi/4$. General solution: $x = \pi/4 + k\pi$ for any integer $k$.

$2u^2 - u - 1 = (2u + 1)(u - 1) = 0 \Rightarrow u = -1/2$ or $u = 1$. $\sin x = -1/2$: 2 solutions ($7\pi/6, 11\pi/6$). $\sin x = 1$: 1 solution ($\pi/2$). Total: $3$.

$\sin x (2\cos x - 1) = 0$. From $\sin x = 0$: $x = 0, \pi$. From $\cos x = \tfrac{1}{2}$: $x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$. Total: $4$.

$\sin x = \cos x \iff \tan x = 1 \iff x = \pi/4 + k\pi$.

$\cos^2 x = \tfrac{1}{2} \Rightarrow \cos x = \pm \tfrac{\sqrt{2}}{2}$. Solutions: $\tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}$.