How to solve $A\sin{\theta} + B\cos{\theta} = C$? I've stumbled upon a equation in the form
$$A\sin{\theta} + B\cos{\theta} = C$$
What would be the steps necessary to solving it?
Thank you.

$$A\sin{\theta} + B\cos{\theta} = C$$
use the identity
$$\cos\theta= \pm \sqrt{1-\sin^2\theta}$$
so
$$A\sin{\theta} \pm B\sqrt{1-\sin^2\theta} = C$$
$$ \pm B\sqrt{1-\sin^2\theta} = C-A\sin{\theta}$$
$$ B^2(1-\sin^2\theta) = (C-A\sin{\theta})^2$$
let $x=\sin{\theta}$
$$ B^2(1-x^2) = (C-Ax)^2$$
$$B^2-B^2x^2=C^2-2ACx+A^2x^2$$
$$(A^2+B^2)x^2-(2AC)x+(C^2-B^2)=0$$
then use the quadratic formula to solve it and then complete the solution to find the $\theta$ values
