Find the set {x in [0, 2pi] mid sin x + i cos | vedclass

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(D) Let $z_1 = \sin x + i \cos 2x$ and $z_2 = \cos x - i \sin 2x$. For $z_1$ and $z_2$ to be conjugates,we must have $z_1 = \overline{z_2}$. This impl

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(D) Let $z_1 = \sin x + i \cos 2x$ and $z_2 = \cos x - i \sin 2x$. For $z_1$ and $z_2$ to be conjugates,we must have $z_1 = \overline{z_2}$. This implies $\sin x + i \cos 2x = \overline{\cos x - i \sin 2x} = \cos x + i \sin 2x$. Equating real and imaginary parts,we get: $1) \sin x = \cos x \implies 	an x = 1 \implies x = \frac{\pi}{4}, \frac{5\pi}{4}$. $2) \cos 2x = \sin 2x \implies 	an 2x = 1 \implies 2x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4} \implies x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}$. Since there is no common value of $x$ that satisfies both equations,the set is empty. Thus,the correct option is $D$.

Find the set $\{x \in [0, 2\pi] \mid \sin x + i \cos 2x \text{ and } \cos x - i \sin 2x \text{ are conjugate to each other}\}$.

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