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

If for certain xx, 3cos⁡x≠2sin⁡x3 \cos x \neq 2 \sin x, then the general solution of, sin⁡2x−cos⁡2x=2−sin⁡2x\sin^2 x - \cos 2x = 2 - \sin 2x, is

A

(2n+1)π2,n∈Z(2n+1)\frac{\pi}{2}, n \in \mathbb{Z}

B

(2n+1)π4,n∈Z(2n+1)\frac{\pi}{4}, n \in \mathbb{Z}

C

nπ+(−1)nπ3,n∈Zn\pi + (-1)^n\frac{\pi}{3}, n \in \mathbb{Z}

D

nπ2+1,n∈Z\frac{n\pi}{2} + 1, n \in \mathbb{Z}

Answer

(2n+1)π2,n∈Z(2n+1)\frac{\pi}{2}, n \in \mathbb{Z}

Explanation

Solution

The equation is sin⁡2x−cos⁡2x=2−sin⁡2x\sin^2 x - \cos 2x = 2 - \sin 2x.

Using the identity cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x, the equation becomes:

sin⁡2x−(1−2sin⁡2x)=2−sin⁡2x\sin^2 x - (1 - 2\sin^2 x) = 2 - \sin 2x 3sin⁡2x−1=2−sin⁡2x3\sin^2 x - 1 = 2 - \sin 2x 3sin⁡2x+sin⁡2x−3=03\sin^2 x + \sin 2x - 3 = 0

Testing the options:

Option A: x=(2n+1)π2,n∈Zx = (2n+1)\frac{\pi}{2}, n \in \mathbb{Z}

When x=π2x = \frac{\pi}{2}: sin⁡π2=1\sin \frac{\pi}{2} = 1, cos⁡π=−1\cos \pi = -1, sin⁡2π2=1\sin^2 \frac{\pi}{2} = 1 sin⁡2x−cos⁡2x=1−(−1)=2\sin^2 x - \cos 2x = 1 - (-1) = 2 sin⁡2x=sin⁡π=0\sin 2x = \sin \pi = 0, so 2−sin⁡2x=22 - \sin 2x = 2 The equality holds.

Verify the condition 3cos⁡x≠2sin⁡x3\cos x \neq 2\sin x:

At x=(2n+1)π2x = \frac{(2n+1)\pi}{2}, cos⁡x=0\cos x = 0 and sin⁡x=±1\sin x = \pm 1. Thus, 3cos⁡x=0≠±2=2sin⁡x3\cos x = 0 \neq \pm 2 = 2\sin x.

Therefore, Option A satisfies the equation and the given condition.