Relation between Circular functions and complex numbers.

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Does the following equation has a solution?

$$1+e^{\sqrt{-x}}= \sin x$$

I came across it while I was solving

$$\int{2\sin x\cos xdx}$$

Which has multiple solutions as $$-\frac{ \cos(2x)}{2}+c_1$$ $$\sin^2(x)+c_2$$ $$-\cos^2(x)+c_3$$

I was checking for value of $c_3$ for which it has a solution equal for which $c_2=0$. I got that equation.

Please help!

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I don't see how you got that equation from the integral you're trying to solve, but working out that equation, use that: $$\sin(x)=\Im(e^{ix})=\frac{e^{ix}-e^{-ix}}{2}$$