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

If $3 \cos x \neq 2 \sin x$, then the general solution of $\sin ^{2} x-\cos 2 x=2-\sin 2 x$ is
  1. $x=n \pi+\frac{\pi}{2}, \quad n \in Z$
  2. $x=n\left(\frac{\pi}{2}\right)+\pi, \quad n \in Z$
  3. $x=n\left(\frac{\pi}{2}\right)+\frac{\pi}{3}, \quad n \in Z$
  4. $x=(2 n+1) \pi, \quad n \in Z$

Solution

Given $\sin ^{2} x-\cos 2 x=2-\sin 2 x$ $\therefore \sin ^{2} x-1+2 \sin ^{2} x=2-\sin 2 x \quad \ldots\left[\because \cos 2 \theta=1-2 \sin ^{2} \theta\right]$ $3 \sin ^{2} x \quad=3-\sin 2 x \Rightarrow \sin 2 x=3\left(1-\sin ^{2} x\right)$ $2 \sin x \cos x=3 \cos ^{2} x \Rightarrow \cos x(3 \cos x-2 \sin x)=0$ $\therefore \cos x=0$ or $3 \cos x=2 \sin x$ But $3 \cos x \neq 2 \sin x$ as per condition given $\therefore \cos \mathrm{x}=0 \Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$

Asked in: MHT CET 2020 (12 Oct Shift 2)

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