Question

3-2cosx-4sinx-cos2x+sin2x=0​

Answer 1

by this u will get answer

Answer 2

Given : $3-2 \cos x-4\sin x - \cos 2x+ \sin 2x =0$

Explanation:

$3-2 \cos x-4\sin x - \cos 2x+ \sin 2x =0\\\\3-\cos 2x-4\sin x+\sin 2x-2 \cos x=0\\\\3- (1-2\sin^2x)-4\sin x+2\sin x\cos x- 2\cos x=0\\\\2\sin^2x-4\sin x+2+2\cos x(\sinx -1)=0\\\\2(\sin^2x-2\sin x+1)+2\cos x(\sin x-1)=0\\\\now\\\\2((\sinx -1)^2+\cos x(\sinx -1))=0\\\\\text{taking common sinx -1 }\\\\(\sin x -1)(\sin x -1 +\cos x)=0\\\\\sin x -1=0$

$\\\\\sin x =1 \\\\sin x = \sin \frac{\pi}{2}\\\thus \\\\x= n \pi +(-1)^n \frac{\pi}{2}\\\\now , \\\\\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x =\frac{1}{\sqrt{2}}\\\\\cos x \cos \frac{\pi}{4}+\sin x \sin \frac{\pi}{4}= \cos \frac{\pi}{4}\\\\\cos (x -\frac{\pi}{4})= \cos \frac{\pi}{4}\\\\thus \\\\x-\frac{\pi}{4}= 2n\pi \pm \frac{\pi}{4}\\\\x = 2n\pi \pm \frac{\pi}{2}$

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