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

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

Solution

Given, $\begin{aligned} & \sin ^2 x-\cos 2 x=2-\sin 2 x \\ & \Rightarrow 1-\cos ^2 x-\left(2 \cos ^2 x-1\right)-2+\sin 2 x=0 \\ & \Rightarrow 1-3 \cos ^2 x+1-2+2 \sin x \cdot \cos x \\ & \Rightarrow \cos x(2 \sin x-3 \cos x)=0 \\ & \Rightarrow \cos x=0 \text { or } 2 \sin x-3 \cos x=0 \end{aligned}$
But $2 \sin x \neq 3 \cos x$ ...[Given] $\begin{aligned} & \Rightarrow \cos x=0 \\ & \Rightarrow x=(2 \mathrm{n}+1) \frac{\pi}{2}, \mathrm{n} \in \mathrm{Z} \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 1)

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