lf $sin x + i cos 2 x, cos x - i sin 2 x$ are conjugate to each other, then $x =$

n π

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(n + 1) \frac{π}{2}

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ϕ

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(n + 1) π

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Solution

The correct option is C $ϕ$
Conjugate of $s i n x + i c o s 2 x = s i n x - i c o s 2 x$ $= c o s x - i s i n 2 x$
which implies
$s i n x = c o s x \Rightarrow t a n x = 1 - - - - - - - - - (1)$ [on comparing real parts]
$c o s 2 x = s i n 2 x \Rightarrow t a n 2 x = 1$ [on comparing imaginary parts]
But $t a n 2 x = 1$
$\Rightarrow \frac{2 t a n x}{1 - t a n^{2} x} = 1$
for this to be true $t a n x \neq 1 - - - - - - - - - - - (2)$
Thus there is no real $x$

Suggest Corrections

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