The values of x for which sin x + i cos 2x an | vedclass

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(B) 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 -

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(B) 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 $z_2 = \cos x - i \sin 2x$. For these to be conjugates,we must have $\sin x = \cos x$ and $\cos 2x = -(-\sin 2x) = \sin 2x$. From $\sin x = \cos x$,we get $	an x = 1$,which implies $x = n\pi + \frac{\pi}{4}$. From $\cos 2x = \sin 2x$,we get $	an 2x = 1$,which implies $2x = m\pi + \frac{\pi}{4}$,or $x = \frac{m\pi}{2} + \frac{\pi}{8}$. Since there is no value of $x$ that satisfies both conditions simultaneously,there is no solution.

The values of $x$ for which $\sin x + i \cos 2x$ and $\cos x - i \sin 2x$ are conjugate to each other are

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