Trigonometric equations practice questions — IB Maths AA

Factor as $(2\sin x+1)(\sin x-1)=0$. $\sin x=1$ gives $x=\pi \mathrm{/}2$; $\sin x=-\frac{1}{2}$ gives $x=7\pi \mathrm{/}6,11\pi \mathrm{/}6$. Total: 3 solutions.

$2u^2-u-1=(2u+1)(u-1)=0\Rightarrow u=-1\mathrm{/}2$ 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: $3$.

$2x=k\pi \Rightarrow x=k\pi \mathrm{/}2$. On $[0,2\pi )$: $0,\pi \mathrm{/}2,\pi ,3\pi \mathrm{/}2$. Four solutions.

Squaring gives $\sin 2x=0$, so $x\in \{0,\frac{\pi }{2},\pi ,\frac{3\pi }{2}\}$. Substituting back into the original eliminates $\pi$ and $\frac{3\pi }{2}$. Solutions: $\{0,\frac{\pi }{2}\}$.

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

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

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

$2{\cos }^{2}x+\cos x-1=0\Rightarrow (2\cos x-1)(\cos x+1)=0$. Roots: $\cos x=\frac{1}{2}$ ($x=\frac{\pi }{3},\frac{5\pi }{3}$) and $\cos x=-1$ ($x=\pi$). Three solutions.

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

Reference angle $\frac{\pi }{4}$. In $[0,2\pi )$ cosine is $\frac{\sqrt{2}}{2}$ at $x=\frac{\pi }{4}$ (Q1) and $x=\frac{7\pi }{4}$ (Q4).