Question

Solve the equation
$\left(viii\right)\sin 3x-\sin x=4{\cos }^{2}x-2$

Answer 1

$\sin 3x-\sin x=4{\cos }^{2}x-2$

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

$\Rightarrow \sin 3x-\sin x=2\cos 2x$

$[\because 1+\cos 2\theta =2{\cos }^2\theta ]$

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

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

$\Rightarrow 2\cos 2x\sin x-2\cos 2x=0$

$\Rightarrow \cos 2x(\sin x-1)=0$

$\Rightarrow \cos 2x=0\text{or}(\sin x-1)=0$

$\Rightarrow 2x=(2n+1)\frac{\pi }{2}\text{or}\sin x=1$

$\Rightarrow x=(2n+1)\frac{\pi }{4}\text{or}\sin x=\sin \frac{\pi }{2}$

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

So, the general solution of $\sin x=\sin \frac{\pi }{2}$ is $x=m\pi +(-1)m\frac{\pi }{2}{,}^m\in Z$

Hence, general solution is
$x=(2n+1)\frac{\pi }{4}\text{and}x=m\pi +{(-1)}^m\frac{\pi }{2},n,m\in Z$