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 question, \[{{\overline{z}}_{1}}={{\overline{z}}_{2}}\] \[\Rightarrow \] \[\sin x-i\cos 2x=\cos x-i\sin 2x\] On comparing the real and imaginary parts, we get \[sin\text{ }x=cos\text{ }x\text{ }and\text{ }cos\text{ }2x=sin\text{ }2x\] \[\Rightarrow \] \[tan\text{ }x=1\]and \[tan\text{ }2x=1\] \[\Rightarrow \] \[x=\frac{\pi }{4},\frac{5\pi }{4},....\] and \[x=\frac{\pi }{8},\frac{5\pi }{8},....\] Hence, there is no common value of\[x\].
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