The complex numbers sin x + icos 2x and cos x | vedclass

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(D) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $a = c$ and $b = -d$.Given $z_1 = \sin x + i\cos 2x$ and $z_2 = \cos x - i

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No value of $x$

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(D) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $a = c$ and $b = -d$.Given $z_1 = \sin x + i\cos 2x$ and $z_2 = \cos x - i\sin 2x$,they are conjugates if:$1) \sin x = \cos x \implies 	an x = 1 \implies x = n\pi + \frac{\pi}{4}$$2) \cos 2x = -(-\sin 2x) = \sin 2x \implies 	an 2x = 1 \implies 2x = m\pi + \frac{\pi}{4} \implies x = \frac{m\pi}{2} + \frac{\pi}{8}$Comparing the two conditions:For $n=0, x = \frac{\pi}{4} = 0.25\pi$.For $m=0, x = \frac{\pi}{8} = 0.125\pi$.For $m=1, x = \frac{\pi}{2} + \frac{\pi}{8} = 0.625\pi$.There is no integer $n$ and $m$ such that $n\pi + \frac{\pi}{4} = \frac{m\pi}{2} + \frac{\pi}{8}$.Thus,there is no value of $x$ for which the complex numbers are conjugate.

The complex numbers $\sin x + i\cos 2x$ and $\cos x - i\sin 2x$ are conjugate to each other for

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