Solving trigonometric equations of the form $a\sin x + b\cos x = c$ Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.
How do you solve this equation without using the method that moves $b\cos x$ to the right side and squaring left and right sides of the equation?
And how does solving $\sqrt{3}\sin x + \cos x = 2$ equal to solving $\sin (x+ \frac{\pi}{6}) = 1$

This is little known, but you can solve the equation without much trigonometry.
WLOG, we can assume that $a^2+b^2=1$ (the coefficients can be normalized). Write
$$a\,S+b\,C=c,\\b\,C=c-aS,\\b^2(1-S^2)=(c-a\,S)^2,\\S^2-2ac\,S+c^2-b^2=0.$$
The solution of the quadratic equation is 
$$S=ac\pm bd$$ where $d=\sqrt{1-c^2}$. 
By symmetry,
$$C=bc\mp ad.$$
If you have enough with the values of the sine and the cosine, you can stop here. Otherwise
$$\theta=\arctan\frac SC.$$

For unnormalized $a,b$, the solution is
$$S=\frac{ac\pm bd}{a^2+b^2},\\
C=\frac{bc\mp ad}{a^2+b^2}$$ and
$$\color{green}{\theta=\arctan\frac{ac\pm bd}{bc\mp ad}}$$
where $d=\sqrt{a^2+b^2-c^2}$. It does not exist when $a^2+b^2<c^2$.

From the above, one can observe that the solution is also given by
$$\theta=\arctan\frac ab\pm\arctan\frac dc$$ but this takes two (costly) arc tangents instead of one.
