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The complex numbers $\sin x+i\cos 2x$ and $\cos x-i\sin 2x$ are conjugate for

$\begin{array}{1 1}(A) x=0 \\ (B)x=\large\frac{n \pi}{2} +\frac{\pi}{2} n \in I \\ (C) x= n \pi n \in I \\(D) No\;value\;of\;x \end{array}$

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$\sin x+i\cos 2x=\cos x-i\sin 2x$
$\Rightarrow \sin x-i\cos 2x=\cos x-i\sin 2x$
$\Rightarrow \sin x=\cos x$ and $\cos 2x=\sin 2x$
$\Rightarrow \tan x=1$ and $\tan 2x=1$
$\Rightarrow$ No value of $x$ is possible
Hence (D) is the correct answer.
answered Apr 9, 2014 by sreemathi.v
 

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