A) \[x=n\pi \]
B) \[x=\left( n+\frac{1}{2} \right)\pi \]
C) \[x=0\]
D) No value of x
Correct Answer: D
Solution :
\[\sin x+i\cos 2x\]and \[\cos x-i\sin 2x\]are conjugate to each other if \[=\frac{|{{z}_{1}}-{{z}_{2}}|}{|\overline{{{z}_{1}}-{{z}_{2}}}|}=\frac{|{{z}_{1}}-{{z}_{2}}|}{|{{z}_{1}}-{{z}_{2}}|}=1\] and \[\cos 2x=\sin 2x\] or \[\tan x=1\]Þ \[x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},......\] ??(i) and \[\tan 2x=1\]Þ \[2x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},........\] or \[x=\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8}\]....... ?..(ii) There exists no value of \[x\]common in (i) and (ii). Therefore there is no value of \[x\] for which the given complex numbers are conjugate.
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