If $3\cos x\ne 2\sin x$, then the general solution of ${\sin }^{2}x-\cos 2x=2-\sin 2x$ is

The correct option is C

$\frac{\left(4\mathrm{n}\pm 1\right)\mathrm{\pi }}{2},n\in Z$

Explanation for the correct option:

Find the general solution of given equation:

Given, ${\sin }^{2}x–\cos 2x=2–\sin 2x$

$\Rightarrow$$1–{\cos }^{2}x–(2{\cos }^{2}x–1)–2+\sin 2x=0$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\Rightarrow$ $1–3{\cos }^{2}x+1–2+2\sin x\cos x=0$

$\Rightarrow$ $\cos x(2\sin x–3\cos x)=0$

$\Rightarrow$ $\cos x=0,2\sin x–3\cos x=0$

$\Rightarrow$ $\cos x=0$ $\left[\because 3\cos x\ne 2\sin x\right]$

$\Rightarrow$ $x=\frac{\mathrm{\pi }}{2}$

$\begin{array}{rcl}\therefore x& =& 2n\pi \pm \frac{\pi }{2}\\ & =& \frac{\mathbf{(}\mathbf{4}\mathbf{n}\mathbf{}\mathbf{\pm }\mathbf{}\mathbf{1}\mathbf{)}\mathbf{\pi }}{\mathbf{2}}\mathbf{,}\mathbf{}\mathit{n}\mathbf{}\mathbf{\in }\mathbf{}\mathit{Z}\end{array}$

Hence, Option ‘C’ is Correct.