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...