AP EAMCET · Maths · Trigonometric Equations
The solution of the equation \([\sin x+\cos x]^{1+\sin 2 x}=2\), where \(-\pi \leq x \leq \pi\) is
- A \(\frac{\pi}{2}\)
- B \(\pi\)
- C \(\frac{\pi}{4}\)
- D \(\frac{3 \pi}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { }(\sin x+\cos x)^{1+\sin 2 x}=2 \text { where }-\pi \leq x \leq \pi \\ & (\sin x+\cos x)^{(\sin x+\cos x)^2}=2\left[\because(\sin x+\cos x)^2\right. \\ & =\sin ^2 x+\cos ^2 x+2 \sin \cos x \\ & =1+\sin 2 x] \\ & \because \sqrt{a^2+b^2} \leq a \sin x+b…
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