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I tried:

$$3 \operatorname*{cis}(\frac{4\pi}{3})\cdot \frac{1}{3} \operatorname*{cis}(\alpha) = \operatorname*{cis}(\frac{4\pi}{3}+\alpha) = \cos(\frac{4\pi}{3}+\alpha) + \sin(\frac{4\pi}{3}+\alpha)i$$

Since this has to be a real number I did:

$$\sin(\frac{4\pi}{3}+\alpha) = 0 \Leftrightarrow \frac{4\pi}{3}+\alpha = 2k\pi \lor \frac{4\pi}{3}+\alpha = \pi+2k\pi \Leftrightarrow \\ \alpha = 2k\pi-\frac{4\pi}{3} \lor \alpha = \pi+2k\pi-\frac{4\pi}{3}$$

But my book states the solution is $\alpha = \frac{2\pi}{3}+k\pi$. What went wrong?

By the way I know theres a shorter way of representing the solution and it goes something like $\alpha = \text{something }+(-1)^k\arcsin(\text{number})$, anyone knows it?

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Note that $$\sin { x } =0\quad \Rightarrow x=k\pi ,\quad k\in \mathbb Z $$ so $$\\ \sin { \left( \frac { 4\pi }{ 3 } +\alpha \right) =0\quad \Rightarrow \frac { 4\pi }{ 3 } +\alpha =k\pi ,\quad \Rightarrow } \alpha =-\frac { 4\pi }{ 3 } +k\pi ,\quad or\quad \alpha =\frac { 2\pi }{ 3 } +k\pi $$

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Nothing went wrong, except you made things more complicated for yourself.

Hint: if I have $x = 2k\pi \vee x = 2k\pi + \pi$, that's the same as saying $x = k\pi$.

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$$\sin (\alpha +\frac {4\pi}{3})=0$$

$$\implies \alpha+\frac {4\pi}{3}=k'\pi\;\;,k'\in\mathbb Z $$ $$\implies \alpha=k'\pi-\frac {4\pi}{3} $$

$$=(k'-2)\pi+\frac {2\pi}{3} $$ $$=k\pi+\frac {2\pi}{3} \;\;,k\in\mathbb Z$$

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