= Solution
Using the <hyperbolic functions> identities $\sin(ix)=i\sinh x$ and $\cos(ix)=\cosh x$, divide by $i$ and put $u=e^x\ne0$. The equation becomes $-u/2-5/(2u)=3$, hence
$$
u^2+6u+5=(u+1)(u+5)=0.
$$
Taking all branches of the <complex logarithm> gives
$$
\boxed{x=(2k+1)\pi i\quad\text{or}\quad x=\ln5+(2k+1)\pi i,\qquad k\in\mathbb Z.}
$$
If $x$ is required to be real, there are \b[no real solutions], since $e^x$ cannot be negative. The complex branches are essential to the unrestricted answer.
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