<s>[INST] Solving $\cos x + 3^{1/2} \sin x = 1$ for $0\leq x \leq 360^{\circ}$ $\cos x + \sqrt3 \sin x = 1$
Not sure what the step would be to get an answer.
 [/INST] If we square both sides of the equation:
$$
  \Big(\cos(x) + \sqrt{3}\sin(x)\Big)^2 = 1^2 \\
  \cos^2(x) + 3\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1  \\
  \cos^2(x) + \sin^2(x) + 2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1  \\
  1 + 2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1  \\
  2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 0  \\
  \sin(x)\Big(\sin(x) + \sqrt{3}\cos(x)\Big) = 0  \\
$$
So either $\sin(x) = 0$ or:
$$
  \sin(x) + \sqrt{3}\cos(x) = 0 \\
  \sin(x) = -\sqrt{3}\cos(x) \\
  \tan(x) = -\sqrt{3} \\
$$
Then for what angles $x$ does $\sin(x) = 0$ or $\tan(x) = -\sqrt{3}$?
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