Q116 / 180 · KCET 2008 · Mathematics

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

  • A
    $n \pi+\frac{\pi}{4}$
  • Correct answer
    B
    $2 \mathrm{n} \pi \pm \frac{\pi}{4}$
  • C
    $n \pi \pm \frac{\pi}{4}$
  • D
    $n \pi-\frac{\pi}{4}$

Solution

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 $)$ $\Rightarrow \quad \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{4}$

Hence, the correct answer is option: 'B'.

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