KCET · Maths · Trigonometric Equations
If \( \cos x=|\sin x| \) then, the general solution is
- A \( x=2 n \pi \pm \frac{\pi}{4}, n \in z \)
- B \( x=(2 n+1) \pi \pm \frac{I}{4}, n \in Z \)
- C \( x=n \pi \pm \frac{\pi}{4}, n \in Z \)
- D \( x=n \pi \pm(-1)^{n} \frac{\pi}{4}, n \in Z \)
Answer & Solution
Correct Answer
(A) \( x=2 n \pi \pm \frac{\pi}{4}, n \in z \)
Step-by-step Solution
Detailed explanation
(A)
\(\cos x=|\sin x|\)
\(\Rightarrow \pm \cos x=\sin x\)
\(\Rightarrow \tan x=\pm 1\)
\(x=n \pi \pm \frac{\pi}{4}, n \in z\), but \(\cos x\) is positive so \(x=2 n \pi \pm \frac{\pi}{4}, n \in z\)
\(\cos x=|\sin x|\)
\(\Rightarrow \pm \cos x=\sin x\)
\(\Rightarrow \tan x=\pm 1\)
\(x=n \pi \pm \frac{\pi}{4}, n \in z\), but \(\cos x\) is positive so \(x=2 n \pi \pm \frac{\pi}{4}, n \in z\)
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