TS EAMCET · Maths · Trigonometric Equations
If \(3 \cos x \neq 2 \sin x\), then the general solution of \(\sin ^2 x-\cos 2 x=2-\sin 2 x\) is
- A \(n \pi+(-1)^n \frac{\pi}{2}, n \in Z\)
- B \(\frac{n \pi}{2}, n \in Z\)
- C \((4 n \pm 1) \frac{\pi}{2}, n \in Z\)
- D \((2 n-1) \pi, n \in Z\)
Answer & Solution
Correct Answer
(C) \((4 n \pm 1) \frac{\pi}{2}, n \in Z\)
Step-by-step Solution
Detailed explanation
\begin{gathered}\sin ^2 x-\cos 2 x=2-\sin 2 x \\ \Rightarrow 1-\cos ^2 x-\left(2 \cos ^2 x-1\right) \\ =2-2 \sin x \cos x \\ \Rightarrow \quad-3 \cos ^2 x+2 \sin x \cos x=0 \\ \Rightarrow \quad \cos x(2 \sin x-3 \cos x)=0 \\ \Rightarrow \quad \cos x=0, \quad(\because 2 \sin x-3…
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