Question

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

• A. $n\pi +(-1{)}^n\frac{\pi }{2},n\in Z$
• B. $\frac{n\pi }{2},n\in Z$
• C. $(4n\pm 1)\frac{\pi }{2},n\in Z$
• D. $(2n-1)\pi ,n\in Z$

Answer 1

Answer: (C)

${\sin }^{2}x-\cos 2x=2-\sin 2x$
$\Rightarrow 1-{\cos }^{2}x-\left(2{\cos }^{2}x-1\right)$
$=2-2\sin x\cos x$
$\Rightarrow -3{\cos }^{2}x+2\sin x\cos x=0$
$\Rightarrow \cos x(2\sin x-3\cos x)=0$
$\Rightarrow \cos x=0,(\because 2\sin x-3\cos x\ne 0)$
$\Rightarrow x=2n\pi \pm \frac{\pi }{2}$
$\Rightarrow x=(4n\pm 1)\frac{\pi }{2}$