$\sin x=2$ ?

$e^{ix}=\cos x+i\sin x \newline e^{-ix}=\cos x-i\sin x \newline e^{ix}-e^{-ix}=2i \sin x \newline e^{ix}-e^{-ix}=2i \times 2 = 4i \newline e^{-ix} = \frac{1}{e^{ix}}$ , so let $t = e^{ix}$ , and $t-\frac{1}{t}=4i \newline t^2-(4i)t-1=0. \newline$ Find the root formula with a quadratic equation in one variable: $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ , inferred $\newline t = \frac{4i \pm \sqrt{-16+4}}{2} \newline = 2i \pm \sqrt{-3} \newline = (2 \pm \sqrt{3})i. \newline$ $e^{ix}=(2 \pm \sqrt{3})i \newline ix = \ln((2 \pm \sqrt{3})i) \newline = \ln(2 \pm \sqrt{3}) + \ln(i). \newline$ It seems to be one step away from success, but what is $\ln(i)$ ? $\newline$ Let $i \theta = \ln(i)$ , inferred $\newline$ $e^{i \theta}=\cos \theta + i \sin \theta \newline i = \cos \theta + i \sin \theta. \newline$ The real number part on the left side of the equation is 0, so the real number part on the right side of the equation is also 0, that is, $\cos \theta = 0. \newline$ The same, $\sin \theta = 1. \newline$ So $\theta = \frac{4k+1}{2} \pi$ (k is an integer). $\newline$ So $ix = \frac{4k+1}{2}i \pi + \ln(2 \pm \sqrt{3}) \newline x = \frac{4k+1}{2} \pi + \frac{1}{\sqrt{-1}} \ln(2 \pm \sqrt{3}) \newline = \frac{4k+1}{2} \pi - i \ln(2 \pm \sqrt{3})$ (k is an integer).

$\begin{aligned} \blue{\sin x} & = 2 & \small \blue{\text{By Euler's formula: }e^{\theta i} = \cos \theta + i \sin \theta} \\ \blue{\frac {e^{ix}-e^{-ix}}{2i}} & = 2 \\ e^{ix}-e^{-ix} & = 4i & \small \blue{\text{Multiply both sides by }e^{ix} \text{ and rearrange.}} \\ e^{2ix} - 4ie^{ix} - 1 & = 0 & \small \blue{\text{Solve the quadratic equation for }e^{ix}} \\ e^{ix} & = \frac {4i \pm \sqrt{-16+4}}2 \\ & = i(2 \pm \sqrt 3) & \small \blue{\text{Take natural log both sides}} \\ ix & = \ln i + \ln (2 \pm \sqrt 3) \\ & = \ln \left(e^{\left(2k + \frac 12\right) \pi i} \right) + \ln (2 \pm \sqrt 3) & \small \blue{\text{where }k \text{ is an integer.}} \\ & = \frac {4k+1}2 \pi i + \ln (2 \pm \sqrt 3) & \small \blue{\text{Divide both sides by }i} \\ \implies x & = \boxed{\frac {4k+1}2\pi - i \ln (2 \pm \sqrt 3)} \end{aligned}$

Reference: Euler's formula