$\sin x=a$

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

Reference: Euler's formula

You must look at the solution I gave to the question sinx=2? so that you can understand the solution. Let $t = e^{ix}$ , so $t - \frac{1}{t} = 2i \times a \newline t^2 - (2ai)t - 1 = 0 \newline t = \frac{2ai \pm \sqrt{-4a^2+4}}{2} \newline = ai \pm \sqrt{1-a^2}. \newline$ Because $a > 1$ , so $\sqrt{1-a^2} < 0$ and $ai \pm \sqrt{1-a^2} = ai \pm \sqrt{a^2-1}i \newline = i(a \pm \sqrt{a^2-1}). \newline e^{ix} = i(a \pm \sqrt{a^2-1}) \newline ix = \ln(i) + \ln(a \pm \sqrt{a^2-1}) \newline \boxed{x = \frac{4k+1}{2} \pi - i\ln(a \pm \sqrt{a^2-1})}$ .