Question

Solve the following equation:
$3-2\cos x-4\sin x-\cos 2x+\sin 2x=0$

Answer 1

$3-2\cos x-4\sin x-\cos 2x+\sin 2x=0$

$\Rightarrow 3-2\cos x-4\sin x-(1-2{\sin }^{2}x)+2\sin x\cos x=0$

$\Rightarrow 2-2\cos x-4\sin x+2{\sin }^{2}x+2\sin x\cos x=0$

$\Rightarrow 2{\sin }^{2}x-4\sin x+2+2\sin x\cos x-2\cos x=0$

$\Rightarrow 2(\sin x-1{)}^2+2\cos x(\sin x-1)=0$

$\Rightarrow 2(\sin x-1)\left[\right(\sin x-1)+\cos x]=0$

$\Rightarrow (\sin x-1)[\sin x+\cos x-1]=0$

$\Rightarrow (\sin x-1)\times \sqrt{2}[\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}]=0$

$\Rightarrow (\sin x-1)[\sin \frac{\pi }{4}\sin x+\cos \frac{\pi }{4}\cos x-\frac{1}{\sqrt{2}}]=0$

$\Rightarrow (\sin x-1)[\cos (x-\frac{\pi }{4})-\frac{1}{\sqrt{2}}]=0$

$\Rightarrow (\sin x-1)=0\text{or}[\cos (x-\frac{\pi }{4})-\frac{1}{\sqrt{2}}]=0$

$\Rightarrow \sin x=1\text{or}\cos (x-\frac{\pi }{4})=\frac{1}{\sqrt{2}}$

$\Rightarrow \sin x=\sin \frac{\pi }{2}\text{or}\cos (x-\frac{\pi }{4})=\cos \frac{\pi }{4}$

We know the general solution of $\sin x=\sin \alpha$ is $x=n\pi +(-1{)}^{n}a,n\in Z$

We know the general solution of $\cos x=\cos \alpha$ is $x=2m\pi \pm \alpha ,m\in Z$

So, $x=n\pi +(-1{)}^n\frac{\pi }{2},n\in Z$ and $(x-\frac{\pi }{4})=2m\pi \pm \frac{\pi }{4}\in Z$

Taking positive sign, we get
$x-\frac{\pi }{4}=2m\pi +\frac{\pi }{4},m\in Z$

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

Taking negative sign, we get
$x-\frac{\pi }{4}=2m\pi -\frac{\pi }{4},m\in Z$

$\Rightarrow x=2m\pi ,m\in Z$

Hence, the general solution is
$x=n\pi +(-1{)}^n\frac{\pi }{2},x=2m\pi$ and $x=2m\pi +\frac{\pi }{2},n,m\in Z$