<s>[INST] how to solve $a\sin x+b\cos x$ Let's solve:
$\sqrt{3}\sin x - \cos x=2$
The left hand side may be expressed as $R\sin(x+ \phi)$
We know that $R=\sqrt{3+1}=2$
We also know that $\tan \phi= \frac{-1}{\sqrt{3}}$
The solution to $\tan \phi=\frac{-1}{\sqrt{3}}$ has many solutions, for example, -30, 150, 330 degrees etc.
Which of these solutions do we accept? Or is it irrelevant which we will accept? Which of these solutions are acceptable?
Thanks! 
 [/INST] $$\frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x=1$$
$$\sin\left(x -\frac{\pi}{6}\right) = 1$$
$$\sin\left(x -\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2}\right)$$
$$ x = n\pi + (-1)^n \frac{\pi}{2} +\frac{\pi}{6}$$
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