Question:
If $3 \cos x \neq 2 \sin x$, then the general solution of $\sin^{2x-\cos 2x=2-\sin 2x$ is
If $3 \cos x \neq 2 \sin x$, then the general solution of $\sin^{2x-\cos 2x=2-\sin 2x$ is
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Always factor out common terms like $\cos x$ first.
Updated On: Apr 10, 2026
- $n\pi+(-1)^{n}\frac{\pi}{2}, n \in Z$
- $\frac{n\pi}{2}, n \in Z$
- $(4n\pm1)\frac{\pi}{2}, n \in Z$
- $(2n-1)\pi, n \in Z$
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The Correct Option is C
Solution and Explanation
Step 1: Use identities
Replace $\cos 2x$ with $2 \cos^2 x - 1$ and $\sin 2x$ with $2 \sin x \cos x$: $\sin^2 x - (2 \cos^2 x - 1) = 2 - 2 \sin x \cos x$.
Step 2: Simplify
$-3 \cos^2 x + 2 \sin x \cos x = 0$ (since $\sin^2 x + \cos^2 x = 1$).
Step 3: Factorize
$\cos x (2 \sin x - 3 \cos x) = 0$.
Step 4: Solve
Since $2 \sin x - 3 \cos x \neq 0$, then $\cos x = 0$. The general solution for $\cos x = 0$ is $x = (2n+1)\frac{\pi}{2}$, which can be written as $(4n \pm 1)\frac{\pi}{2}$.
Final Answer: (C)
Replace $\cos 2x$ with $2 \cos^2 x - 1$ and $\sin 2x$ with $2 \sin x \cos x$: $\sin^2 x - (2 \cos^2 x - 1) = 2 - 2 \sin x \cos x$.
Step 2: Simplify
$-3 \cos^2 x + 2 \sin x \cos x = 0$ (since $\sin^2 x + \cos^2 x = 1$).
Step 3: Factorize
$\cos x (2 \sin x - 3 \cos x) = 0$.
Step 4: Solve
Since $2 \sin x - 3 \cos x \neq 0$, then $\cos x = 0$. The general solution for $\cos x = 0$ is $x = (2n+1)\frac{\pi}{2}$, which can be written as $(4n \pm 1)\frac{\pi}{2}$.
Final Answer: (C)
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