Question

Solve the following equation:
-sin^2x=2cosx-2

Answer 1

`-\sin^2x=2\cos x-2\ \ |\text{use:}\sin^2\alpha+\cos^2\alpha=1\to \sin^2\alpha=1-\cos^2\alpha\\\\-(1-\cos^2x)=2\cos x-2\\\\-1+\cos^2x=2\cos x-2\ \ \ |-2\cos x\\\\\cos^2x-2\cos x-1=-2\ \ \ \ |+2\\\\\cos^2x-2\cos x+1=0\\\\\cos^2x-\cos x-\cos x+1=0\\\\\cos x(\cos x-1)-1(\cos x-1)=0\\\\(\cos x-1)(\cos x-1)=0\\\\(\cos x-1)^2=0\iff\cos x-1=0\ \ \ |+1\\\\\cos x=1\\\\\boxed{x=2k\pi},\ k\in\mathbb{Z}`

Answer 2

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