Question

Solve the equation
$\left(vii\right)\cos x+\sin x=\cos 2x+\sin 2x$

Answer 1

$\cos x+\sin x=\cos 2x+\sin 2x$

$\Rightarrow \cos x-\cos 2x=\sin 2x-\sin x$

Using $\cos C-\cos D=-2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$
And
$\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

$\Rightarrow -2\sin \left(\frac{x+2x}{2}\right)\sin \left(\frac{x-2x}{2}\right)=2\cos \left(\frac{x+2x}{2}\right)\sin \left(\frac{2x-x}{2}\right)$

$\Rightarrow -\sin \left(\frac{3x}{2}\right)\sin (-\frac{x}{2})=\cos \left(\frac{3x}{2}\right)\sin \left(\frac{x}{2}\right)$

$\Rightarrow \sin \left(\frac{3x}{2}\right)\sin \left(\frac{x}{2}\right)-\cos \left(\frac{3x}{2}\right)\sin \left(\frac{x}{2}\right)=0$

$\Rightarrow \sin \left(\frac{x}{2}\right)[\sin \left(\frac{3x}{2}\right)-\cos \left(\frac{3x}{2}\right)]=0$

$\Rightarrow \sin \left(\frac{x}{2}\right)=0\text{or}[\sin \left(\frac{3x}{2}\right)-\cos \left(\frac{3x}{2}\right)]=0$

$\Rightarrow \frac{x}{2}=n\pi ,n\in Z\text{or}\sin \left(\frac{3x}{2}\right)=\cos \left(\frac{3x}{2}\right)$

$\Rightarrow x=2n\pi ,n\in Z\text{or}\tan \left(\frac{3x}{2}\right)=1$

$\Rightarrow x=2n\pi ,n\in Z\text{or}\tan \left(\frac{3x}{2}\right)=\tan \frac{\pi }{4}$

We know the general solution of $\tan x=\tan \alpha$ is $x=m\pi +\alpha ,m\in Z$

So, the general solution of $\tan \left(\frac{3x}{2}\right)=\tan \frac{\pi }{4}$ is $\frac{3x}{2}=m\pi +\frac{\pi }{4},m\in Z$

$\Rightarrow x=\frac{2m\pi }{3}+\frac{\pi }{6},m\in Z$

Hence, the general solution is
$x=2n\pi \text{and}x=\frac{2m\pi }{3}+\frac{\pi }{6},n,m\in Z$