The general solution of |sin x|=cos x is (whe | vedclass

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(B) Given,$|\sin x| = \cos x$. Since $|\sin x| \ge 0$,we must have $\cos x \ge 0$. Squaring both sides,we get $\sin^2 x = \cos^2 x$. Using the identit

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$2 n \pi \pm \frac{\pi}{4}$

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(B) Given,$|\sin x| = \cos x$. Since $|\sin x| \ge 0$,we must have $\cos x \ge 0$. Squaring both sides,we get $\sin^2 x = \cos^2 x$. Using the identity $\sin^2 x = 1 - \cos^2 x$,we have $1 - \cos^2 x = \cos^2 x$. This simplifies to $2 \cos^2 x = 1$,or $\cos^2 x = \frac{1}{2}$. Taking the square root,$\cos x = \pm \frac{1}{\sqrt{2}}$. Since $\cos x \ge 0$,we discard the negative value,so $\cos x = \frac{1}{\sqrt{2}}$. The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$. Here,$\cos x = \cos(\frac{\pi}{4})$,so $x = 2n\pi \pm \frac{\pi}{4}$.

The general solution of $|\sin x|=\cos x$ is (when $n \in I$) given by

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