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 $\sin (x+ \frac{\pi}{6}) = 1$

Riffing on @Yves' "little known" solutions ... 

The above trigonograph shows a scenario with $a^2 + b^2 = c^2 + d^2$, for $d \geq 0$, and we see that
$$\theta = \operatorname{atan}\frac{a}{b} + \operatorname{atan}\frac{d}{c} \tag{1}$$
(If the "$a$" triangle were taller than the "$b$" triangle, the "$+$" would become "$-$". Effectively, we can take $d$ to be negative to get the "other" solution.) 
Observe that both $c$ and $d$ are expressible in terms of $a$, $b$, $\theta$:
$$\begin{align}
a \sin\theta + b \cos\theta &= c \\
b \sin\theta - a\cos\theta &= d \quad\text{(could be negative)}
\end{align}$$
Solving that system for $\sin\theta$ and $\cos\theta$ gives
$$\left.\begin{align}
\sin\theta &= \frac{ac+bd}{a^2+b^2} \\[6pt]
\cos\theta &= \frac{bc-ad}{a^2+b^2}
\end{align}\quad\right\rbrace\quad\to\quad \tan\theta = \frac{ac+bd}{bc-ad} \tag{2}$$
We can arrive at $(2)$ in a slightly-more-geometric manner by noting
$$c d = (a\sin\theta + b \cos\theta)d = c( b\sin\theta - a \cos\theta ) \;\to\; ( b c - a d)\sin\theta = \left( a c + b d \right)\cos\theta \;\to\; (2) $$
where each term in the expanded form of the first equation can be viewed as the area of a rectangular region in the trigonograph. (For instance, $b c \sin\theta$ is the area of the entire figure.)
