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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The solution of the equation $[\sin x+\cos x]^{1+\sin 2x}=2,-\pi\leq x\leq \pi$ is

$\begin{array}{1 1}(a)\;\large\frac{\pi}{2}&(b)\;\large\frac{\pi}{6}\\(c)\;\large\frac{\pi}{4}&(d)\;0\end{array}$

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1 Answer

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Let $I=[\sin x+\cos x]^{\large 1+\sin 2x}$
$\qquad=[\sqrt 2\sin(\large\frac{\pi}{4}$$+x)]^{1+\sin 2x}$
$x=\large\frac{\pi}{4}$
$I=[\sqrt 2\sin(\large\frac{\pi}{4}+\frac{\pi}{4})]^{1+\sin \large\frac{2\pi}{4}}$
$\;\;=(\sqrt 2)^2$
$\;\;=2$
Hence (c) is the correct answer.
answered Nov 15, 2013 by sreemathi.v
 

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