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KCET · Maths · Trigonometric Equations

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

  1. A \(n \pi+\frac{\pi}{4}\)
  2. B \(2 \mathrm{n} \pi \pm \frac{\pi}{4}\)
  3. C \(n \pi \pm \frac{\pi}{4}\)
  4. D \(n \pi-\frac{\pi}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2 \mathrm{n} \pi \pm \frac{\pi}{4}\)

Step-by-step Solution

Detailed explanation

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