trigonometry - Finding a solution to sin(x) = -5 using Euler's Formula

I have just learned about Euler's formula and I am attempting to find a solution to $\sin(\theta)=-5$. However, I am not entirely sure how to precede.

Thank You So Much!

Answer

HINT (you only need to simplify the final answer):

$$\sin(x)=-5\Longleftrightarrow$$
$$\frac{1}{2}\left(ie^{-ix}-ie^{ix}\right)=-5\Longleftrightarrow$$
$$ie^{-ix}-ie^{ix}=-10\Longleftrightarrow$$

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Substitute $y=-ie^{ix}$:

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$$\frac{1}{(0-i)e^{(0+i)x}}+(0-i)e^{(0+i)x}=-10\Longleftrightarrow$$
$$y+\frac{1}{y}=-10\Longleftrightarrow$$
$$\frac{y^2+1}{y}=-10\Longleftrightarrow$$
$$y^2+1=-10\Longleftrightarrow$$
$$y^2+10y+1=0\Longleftrightarrow$$
$$y^2+10y=-1\Longleftrightarrow$$

$$y^2+10y+25=24\Longleftrightarrow$$
$$(y+5)^2=24\Longleftrightarrow$$
$$y+5=\pm\sqrt{24}\Longleftrightarrow$$
$$y+5=\pm 2\sqrt{6}\Longleftrightarrow$$
$$y=\pm 2\sqrt{6}-5\Longleftrightarrow$$
$$-ie^{ix}=\pm 2\sqrt{6}-5\Longleftrightarrow$$
$$e^{ix}=\frac{\pm 2\sqrt{6}-5}{-i}\Longleftrightarrow$$
$$ix=\ln\left(\frac{\pm 2\sqrt{6}-5}{-i}\right)+2i\pi n\Longleftrightarrow$$
$$x=\frac{\ln\left(\frac{\pm 2\sqrt{6}-5}{-i}\right)+2i\pi n}{i}$$