KCET · Maths · Trigonometric Equations
The general solution of \(|\sin x|=\cos x\) is (when \(\mathrm{n} \in \mathrm{I}\) ) given by
- A \(n \pi+\frac{\pi}{4}\)
- B \(2 \mathrm{n} \pi \pm \frac{\pi}{4}\)
- C \(n \pi \pm \frac{\pi}{4}\)
- D \(n \pi-\frac{\pi}{4}\)
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 \()\)
\(\Rightarrow \quad \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{4}\)
\[
\begin{array}{lr}
\therefore & \sin ^{2} \mathrm{x}=\cos ^{2} \mathrm{x} \\
\Rightarrow & 1-\cos ^{2} \mathrm{x}=\cos ^{2} \mathrm{x} \\
\Rightarrow & 2 \cos ^{2} \mathrm{x}=1
\end{array}
\]
\(\Rightarrow \quad \cos x=\pm \frac{1}{\sqrt{2}}\)
\(\Rightarrow \quad \cos x=+\frac{1}{\sqrt{2}} \quad(\because \cos x\) cannot be negative \()\)
\(\Rightarrow \quad \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{4}\)
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