Solving equations with exponentials and trig algebraically Is it possible to algebraically solve an equation of the following form?

$A\sin(x)+Be^x=C$

If so, how?
 A: This kind of equations, which mich trigonometric and non trigonometric functions do not show explicit solutions; for example $x=\cos(x)$ can not be solved analytically.
So, the only way of solving it should be vased on numerical methods. One of the simplest method for solving $f(x)=0$ is Newton which, starting from a reasonable guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
For illustration purposes, let me choose $A=10$, $B=1$, $C=123$. So $$f(x)=10\sin(x)+e^x-123$$ $$f'(x)=10\cos(x)+e^x$$ Let us select, for this specific case $x_0=4$ and apply the method; the successive iterates will then be $5.58067$, $5.08306$, $4.90751$, $4.88935$, $4.88918$ which is the solution for six significant figures.
Starting at $x_0=6$ would have been giving a different type of convergence with the following iterates : $5.32781$, $4.97448$, $4.89289$, $4.88918$.
May be, one of these days, you will ask more about reasonable.
A: As others say, there is little chance that a closed-form solution exists.
For convenience, we can take the logarithm of the unknown to let a line equation appear, and rewrite:
$$y=\sin(\ln x')=ax'+b.$$
The LHS expression is a curve confined in the band $x'>0,|y|\le1$, and you can ignore the straight lines that do not cross it.
Most straight lines that cross the band will have an intersection with the curve in some vicinity of $x'=-b/a$ (on the horizontal axis). This can be used as a starting value for Newton's iterations.
Also, straight lines that go through $x'=0,|y|<1$ (i.e. $|b|<1$) will cross the curve at infinitely many points with $x'$ smaller and smaller, so that $\sin(\ln x')\approx b$, i.e. $x'=\exp(2k\pi+\arcsin b)$ or $x'=\exp((2k+1)\pi-\arcsin b)$, with negative values of $k$.

