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\sin x + i \cos 2x and \cos x - i \sin 2x are conjugate to each other for:
A. x = n\pi

B. 
 x = \left ( n+\frac{1}{2} \right )\frac{\pi}{2}
C. x = 0
D. No value of x

Answers (1)

The answer is the option (c).

\text{let }z=\sin x+i \cos 2 x \quad \bar{z}=\sin x-i \cos 2 x

But \: \: we \: \: are\: \: given\: \: that\: \: \bar{z}=\cos x-i \sin 2 x

\sin x-\operatorname{icos} 2 x=\cos x-i \sin 2x

Comparing the real and imaginary part, we get  \sin \mathrm{x}=\cos \mathrm{x}$ \: \: and\: \: $\cos 2 \mathrm{x}=\sin 2 \mathrm{x}

\tan \mathrm{x}=1 \text { and } \tan 2 \mathrm{x}=1

\tan 2 \mathrm{x}= \frac{2\tan x }{1-\tan^2 x}=1

The above value is not satisfied by tan x = 1. Hence no value of x is possible.

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