Solving $ \sin x + \sqrt 3 \cos x = 1 $ - is my solution correct? I have an equation that I'm trying to solve:
$$ \sin x + \sqrt 3 \cos x = 1 $$
After pondering for a while and trying different things out, this chain of steps is what I ended up with:
$$ \sin x + \sqrt 3 \cos x = 1 $$
$$ \sin x = 1 - \sqrt 3 \cos x $$
$$ \left(\sin x \right)^2 = \left(1- \sqrt 3 \cos x\right)^2 $$
$$ \sin^2 x = 1 - 2 \sqrt 3 \cos x + 3 \cos^2 x $$
$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = 1 - \sin^2 x $$
$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = \cos^2 x $$
$$ 2 \sqrt 3 \cos x = \cos^2 x + 3 \cos^2 x $$
$$ 4 \cos^2 x = 2 \sqrt 3 \cos x $$
$$ \frac{4 \cos^2 x}{\cos x} = 2 \sqrt 3 $$
$$ 4 \cos x = 2 \sqrt 3 $$
$$ \cos x = \frac{2 \sqrt 3}{4} $$
$$ \cos x = \frac{\sqrt 3}{2} $$
The fraction $ \frac{\sqrt 3}{2} $ can be rewritten as $ \cos \left(\pm \frac{\pi}{6}\right) $, so my solutions are:
$$ \cos x = \cos \left(\frac{\pi}{6}\right) \quad \text{or} \quad \cos x = \cos \left(-\frac{\pi}{6}\right) $$
$$ x = \frac{\pi}{6} + 2\pi n \quad \text{or} \quad x = -\frac{\pi}{6} + 2\pi n $$
Since I earlier on exponentiated both sides I have to check my solutions:
$$ x = \frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(\frac{\pi}{6} + 2\pi\right) = 2 \not = \text{RHS} $$
$$ x = -\frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(-\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(-\frac{\pi}{6} + 2\pi\right) = 1 = \text{RHS} $$
Leaving $ x = -\frac{\pi}{6} + 2\pi n $ as the answer since its positive counterpart was not equal to $ 1 $.
$$ \text{Answer:} \: x = -\frac{\pi}{6} + 2\pi n $$
Have I done anything wrong or does this look good? I haven't really done this before so I feel uncertain not just about the solution, but also my steps and notation...

There's a nice trick:
$$\sin x + \sqrt 3 \cos x = 1 \\\\ = 2 \left(\frac{1}{2}\sin x + \frac{\sqrt 3}{2} \cos x\right) \\\\ = 2\left(\cos\left(\frac{\pi}{3} + 2k\pi\right)\sin x + \sin\left(\frac{\pi}{3} + 2k\pi\right)\cos x \right)\\\\= 2\sin\left(x + \frac{\pi}{3} + 2k\pi\right) = 1.$$
When is $$\sin\left(x + \frac{\pi}{3} + 2k\pi\right) = 1/2$$
true?
