solve $\sin z + 2i \cos z = 1$ I'm trying to solve:
$\sin z + 2i \cos z = 1$

I used the formulas:
$\cos z = (e^{iz}+e^{-iz})/{2}$ and $\sin z = (e^{iz}-e^{-iz})/{2i}$

My argument goes like this
$$\begin{split}
1&=(e^{iz}-e^{-iz})/2i+2i(e^{iz}+e^{-iz})/{2}\\
&=(e^{iz}-e^{-iz})/2i+i(e^{iz}+e^{-iz})\\
&=((e^{iz}-e^{-iz})+2i^2(e^{iz}+e^{-iz}))/2i\\
&=(e^{iz}-e^{-iz}-2e^{iz}-2e^{-iz})/2i\\
&=(-e^{iz}-3e^{-iz})/2i\\
\end{split}$$
which is where I get stuck.
is there a way to separate the $z$?
 A: Let $u = e^{iz}$.  Then:
$$\frac{-u-\frac{3}{u}}{2i}=1$$
Multiplying by $2iu$ to eliminate the fractions, and rearranging into a quadratic, gives:
$$u^2 + 2iu + 3 = 0$$
$$u = \frac{-2i \pm \sqrt{(2i)^2-4(1)(3)}}{2} = -i \pm 2i$$
$$u \in \{i, -3i\} =  \{ 1 \operatorname{cis}(\frac{\pi}{2}), 3 \operatorname{cis}(\frac{-\pi}{2}) \}$$
$$\log u = iz \in \{i(\frac{\pi}{2} + 2\pi k), \log 3 + i(\frac{-\pi}{2} + 2\pi k) \}, k \in \mathbb{Z}$$
$$z \in \{\frac{\pi}{2} + 2\pi k, \frac{-\pi}{2} + 2\pi k - i\log 3 \}, k \in \mathbb{Z}$$
A: Turning the comments into an answer:
Rewrite your equation as
$\exp(iz)+2i+3\exp(-iz)=0$
and multiply by $\exp(iz) to get
$[\exp(iz)]^2+2i\exp(iz)+3=0.$
Now solve this as a quadratic equation for $\exp(iz)$. Despite the complex coefficient the discriminant $\Delta$ is real, so the quadratic formula can be applied with no real issues:
$\Delta=(2i)^2-(4×1×3)=-16, \sqrt\Delta=4i$
$\exp(iz)=[-2i\pm(4i)]/2\in\{i,-3i\}.$
And get values of $z$ for each of the two roots indicated.
