A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{8}\]
C) \[\frac{3\pi }{4}\]
D) None of these
Correct Answer: D
Solution :
Let \[{{z}_{1}}=\sin x+i\cos 2x\] and \[{{z}_{2}}=\cos x-i\sin 2x\] According to the given condition, \[{{\bar{z}}_{1}}={{z}_{2}}\] \[\Rightarrow \] \[\sin x-i\cos 2x=\cos x-i\sin 2x\] \[\Rightarrow \] \[\sin x=\cos x\]and\[\cos 2x=\sin 2x\] \[\Rightarrow \] \[\tan x=1\]and\[\tan 2x=1\] \[\Rightarrow \] \[x=\frac{\pi }{4},\,\,\frac{5\pi }{4},...\]and\[x=\frac{\pi }{8},\,\,\frac{5\pi }{8},\,\,...\] Hence, there exists no value of \[x\] common.
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