Solution for $(a\cdot x+b)\cdot \sin(x)+c\cdot \cos(x)=0$? As part of an engineering problem, I've been trying for a generic solution for $(a\cdot x+b)\cdot \sin(x)+c\cdot \cos(x)=0$
Here's something I tried:
$(a⋅x+b)⋅\tan(x)+c=0$, when $\cos(x)\neq 0$
$$\tan(x)=\frac{-c}{a⋅x+b}$$
It sort of looking like this, when, say, $a=2, b=3, c=5$:
Any chance that can be solved without using numerical methods?
EDIT: I was looking for solutions in the form of $x \approx f(a,b,c) $, preferably finding solution in any range, because $a$, $b$, and $c$ can take wierd values. It's used for a program where it will not be trivial to solve problem procedurally.
 A: There's no closed form solution for that equation, so numerical methods are required. 
That said, you can come up with estimates for the solutions towards $\pm \infty$. 
For instance, as $n\rightarrow +\infty$, if you write the solutions as $x_n=n\pi +  u_n$, with $u_n\in (-\frac \pi 2, \frac \pi 2)$, then
$$\tan x_n = \tan u_n = \frac {-c}{ax_n+b}\tag{1}=-\frac c {an\pi + a u_n + b}$$
Because the right-hand side of that equation is equivalent to $-\frac c {a\pi n}$, we must have that $\tan u_n\rightarrow 0$, which means that $u_n\rightarrow 0$.
Plugging that information back into (1) yields
$$u_n=-\frac c {a\pi n} + v_n$$ where $v_n =  o(\frac 1 n)$.
And you can now plug this back, again, into (1), and find an estimate for $v_n$. And you keep doing that ad lib to find higher order estimates in the solution
$$x_n = n\pi -\frac c {a\pi n} + ....$$
A: Besides numerical solutions for particular parameter values, you could try series solutions.  Thus when $c$ is small compared to $a$ and $b$, the solution near $x=0$ is
$$ -\frac{b}{a} + \cot\left(\frac{b}{a}\right) \frac{c}{a} + \cot\left(\frac{b}{a}\right) \csc^2 \left(\frac{b}{a}\right) \frac{c^2}{a^2}
+  \left(\cot^3\left(\frac{b}{a}\right) + \cot\left(\frac{b}{a}\right) \sec^4  \left(\frac{b}{a}\right)\right) \frac{c^3}{a^3} + \ldots $$
