Trigonometric Equations practice questions — A-Level Maths

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

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

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

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

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

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

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

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

$\sin x=\cos x⟺\tan x=1⟺x=\pi \mathrm{/}4+k\pi$.

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