Question

Find all solutions to the equation in the interval [0,2n) cos x = sin 2x

Answer 1

`\huge\boxed{x\in\left\{(\pi)/(6);\ (\pi)/(2)\ (3\pi)/(2);\ (5\pi)/(6)\right\}}`

Explanation:

`\cos x=\sin2x\qquad|\text{use}\ \sin\theta=2\sin\theta\cos\theta\\\\\cos x=2\sin x\cos x\qquad|\text{subtract}\ \cos x\ \text{from both sides}\\\\0=2\sin x\cos x-\cos x\\\\2\sin x\cos x-\cos x=0\qquad|\text{distribute}\\\\\cos x(2\sin x-1)=0\iff\underbrace{\cos x=0}_((1))\ \vee\ \underbrace{2\sin x-1=0}_((2))`

`(1)\\\cos x=0\Rightarrow x=(\pi)/(2)+k\pi;\ k\in\mathbb{Z}`

`(2)\\2\sin x-1=0\qquad|\text{add 1 to both sides}\\\\2\sin x=1\qquad|\text{divide sides by 2}\\\\\sin x=(1)/(2)\Rightarrow x=(\pi)/(6)+2k\pi\ \vee\ x=(5\pi)/(6)+2k\pi`

`\text{From}\ (1)\ \text{and}\ (2)\ \text{we have}\\\\x=(\pi)/(2)+k\pi\ \vee\ x=(\pi)/(6)+2k\pi\ \vee\ x=(5\pi)/(6)+2k\pi\\\\\text{We have the interval}\ x\in[0;2\pi).\ \text{Therefore the solution is:}\\\\x\in\left\{(\pi)/(6);\ (\pi)/(2)\ (3\pi)/(2);\ (5\pi)/(6)\right\}`