The solution of the equation cos^2 x - 2cos x | vedclass

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(C) Given equation: $\cos^2 x - 2\cos x = 4\sin x - \sin 2x$Using $\sin 2x = 2\sin x \cos x$,we get:$\cos^2 x - 2\cos x = 4\sin x - 2\sin x \cos x$$\c

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$\pi + 	an^{-1}\left(-\frac{1}{2}ight)$

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(C) Given equation: $\cos^2 x - 2\cos x = 4\sin x - \sin 2x$Using $\sin 2x = 2\sin x \cos x$,we get:$\cos^2 x - 2\cos x = 4\sin x - 2\sin x \cos x$$\cos x(\cos x - 2) = 2\sin x(2 - \cos x)$$\cos x(\cos x - 2) + 2\sin x(\cos x - 2) = 0$$(\cos x - 2)(\cos x + 2\sin x) = 0$Since $\cos x - 2 
eq 0$ for any real $x$,we must have $\cos x + 2\sin x = 0$.$	an x = -\frac{1}{2}$Since $0 \le x \le \pi$,and $	an x$ is negative in the second quadrant,the solution is $x = \pi + 	an^{-1}\left(-\frac{1}{2}ight)$.

The solution of the equation $\cos^2 x - 2\cos x = 4\sin x - \sin 2x$ for $0 \le x \le \pi$ is

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