Solution for the trignometric equation I am looking for a solution for an equation of the form :
$ax - \sin(bx) + c = 0$. Without the constant term $c$, I can easily take a derivative to get the solution. But how do I take into account the constant?
This gives me the answer, but how do I find the solution?
 A: Are you looking for an analytic solution? You can approximate it with  Newton's method, for example.
A: That is a trascendental equation. In general, you can only solve it using numerical methods. You can find an analytic solution for some specific cases, though.
A: 
$x=-\dfrac ca+\dfrac1a\cdot\sin\bigg(-\dfrac{bc}a+\dfrac ba\cdot\sin\bigg(-\dfrac{bc}a+\dfrac ba\cdot\sin\bigg(-\dfrac{bc}a+\dfrac ba\cdot\sin\bigg(\cdots\bigg)\bigg)\bigg)\bigg)\bigg)\bigg)$

A: When you face equations which mix polynomial and trigonometric terms, you cannot expect any analytical solution (remember that there is no closed form solution to the equation $x=\cos(x)$) and only numerical methods can be used for getting the solution.
Considering the equation 
$$f(x)=ax - \sin(bx) + c $$ yo can notice that it is bounded by $ax+(c-1)$ and $ax+(c+1)$ and then the solution is somewhere between $x_1=-\frac {c-1}a$ and $x_2=-\frac {c+1}a$ which means that $x_0=-\frac {c}a$ (as Lucian wrote) is probably a good starting point for Newton method. The iterates will then be given by $$x_{n+1}=\frac{b\, x_n \cos (b x_n)-\sin (b x_n)+c}{b \cos (b x_n)-a}$$
But we can generate a better estimate computing $f(x_1)$ and $f(x_2)$, establishing the equation of the straight line joining these points and, from there, computing the $x$ intercept. This will give $$x_0=\frac{c-c \sin \left(\frac{b}{a}\right) \cos \left(\frac{b
   c}{a}\right)+\cos \left(\frac{b}{a}\right) \sin \left(\frac{b
   c}{a}\right)}{a \left(\sin \left(\frac{b}{a}\right) \cos \left(\frac{b
   c}{a}\right)-1\right)}$$
You should notice that the example you added to the post leads to an approximate solution $x\approx 11.0000$ just because $11 \approx 3.50141 \pi$  which makes $$\sin(11)\approx \sin(3.50141 \pi)=-\sin(0.50141 \pi)=-\sin(\frac \pi 2+0.00141\pi)=-\cos(0.00141\pi)$$ which is $\approx -0.99999$. If you apply Newton method to this case ($a=1,b=-1,c=-10$),   the successive iterates will be $$x_0=10.999966682789284969$$ $$x_1=10.999990249931731937$$ $$x_2=10.999990249655250451$$ which is the solution for twenty significant figures.
