I have been working on a collision detection algorithm and the whole thing got down to one simple equation: $$ ax + \sin(bx) + c = 0 $$ Which is basically problem of finding intersection between a line and a sine function.
Is there and way to solve this? Maybe expansion into the Taylor's series? Wolfram Alpha does not provide any general answers so I am guessing it is a complicated problem.
Yes, this can be solved, rather easily. WLOG, $a>0$.
Consider the function $$y(x)=ax+\sin bx+c$$ and find its extrema, located at the roots of the derivative,
$$y'(x)=a+b\cos bx=0.$$
Unless $a>|b|$, the solutions are given by
$$x_{\pm k}=\pm\frac1b\arccos\left(-\frac ab\right)+2k\pi.$$
Now plug these values in the function to get the extremal ordinates, $y_{\pm k}=y(x_{\pm k})$. It is guaranteed that the interval between two successive values contains a root if and only if the values at endpoints differ in sign.
Also note that
$$ax-1+c\le ax+\sin bx+c\le ax+1+c$$ so that roots can only exist when
$$-\frac{1+c}a\le x\le \frac{1-c}a.$$
Now, find all $x_{\pm k}$ that satisfy the above bracketing, and test all intervals of which they are an endpoint for a change of sign of $y_{\pm k}$.
When a change of sign is found, the root can be refined to arbitrary accuracy by the regula falsi algorithm, which is guaranteed to converge. If you are just interested in the existence of roots, you needn't use it.
In the given example, the black and magenta lines delimit the interval outside of which no root is possible. This interval contains six extrema (roots of the green sinusoid), hence there are seven intervals to be tried. All seven intervals feature a change of sign, hence there are exactly seven real roots.