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

I'm assuming $ab\neq 0$, since otherwise trivial. $a\sin x+b\cos x-c=0$   
$$\iff a\left(2\sin\frac{x}{2}\cos\frac{x}{2}\right)+b\left(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\right)-c\left(\sin^2\frac{x}{2} +\cos^2\frac{x}{2}\right)=0$$  


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*Assume $\cos\frac{x}{2}\neq 0$. Then 


$$\stackrel{:\cos^2 \frac{x}{2}\neq 0}\iff (b+c)\tan^2\frac{x}{2}-2a\tan\frac{x}{2}+(c-b)=0$$   
If $b+c\neq 0$, then
$$\iff \tan\frac{x}{2}=\frac{a\pm\sqrt{a^2+b^2-c^2}}{b+c}$$   
$$\iff x=2\left(\arctan\left(\frac{a\pm\sqrt{a^2+b^2-c^2}}{b+c}\right)+n\pi\right),\, n\in\Bbb Z$$   
Real solutions exist iff $a^2+b^2\ge c^2$.   
If $b+c=0$, then
$$\iff \tan\frac{x}{2}=\frac{c-b}{2a}\iff x=2\left(\arctan\frac{c-b}{2a}+n\pi\right),\, n\in\Bbb Z$$  


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*Assume $\cos\frac{x}{2}=0$. Equality holds iff $b+c=0$.

