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The general solution of $|\sin x|=\cos x$ is (when $\mathrm{n} \in \mathrm{I}$ ) given by
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Verified Answer
The correct answer is:
$2 \mathrm{n} \pi \pm \frac{\pi}{4}$
Given, $|\sin x|=\cos x$
$$
\begin{array}{lr}
\therefore & \sin ^{2} \mathrm{x}=\cos ^{2} \mathrm{x} \\
\Rightarrow & 1-\cos ^{2} \mathrm{x}=\cos ^{2} \mathrm{x} \\
\Rightarrow & 2 \cos ^{2} \mathrm{x}=1
\end{array}
$$
$\Rightarrow \quad \cos x=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \quad \cos x=+\frac{1}{\sqrt{2}} \quad(\because \cos x$ cannot be negative $)$
$\Rightarrow \quad \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{4}$
$$
\begin{array}{lr}
\therefore & \sin ^{2} \mathrm{x}=\cos ^{2} \mathrm{x} \\
\Rightarrow & 1-\cos ^{2} \mathrm{x}=\cos ^{2} \mathrm{x} \\
\Rightarrow & 2 \cos ^{2} \mathrm{x}=1
\end{array}
$$
$\Rightarrow \quad \cos x=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \quad \cos x=+\frac{1}{\sqrt{2}} \quad(\because \cos x$ cannot be negative $)$
$\Rightarrow \quad \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{4}$
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