<s>[INST] How to solve$\frac {1}{\sin {x}} + \frac {\sqrt {3}}{\cos {x}} = 4$ $$\frac {1}{\sin {x}} + \frac {\sqrt {3}}{\cos {x}} = 4$$
Can you help me solve this?
 [/INST] We may assume $x\not\in \frac{\pi}{2}\mathbb{Z}$, hence your equation is equivalent to:
$$ \cos(x)+\sqrt{3}\sin(x) = 4\sin(x)\cos(x) = 2\sin(2x) $$
or to:
$$ \sin\left(x+\frac{\pi}{6}\right) = \sin(2x) $$
whose fundamental solutions are given by $2x=x+\frac{\pi}{6}$, leading to $x=\frac{\pi}{6}$, and by $x+\frac{\pi}{6}=(2k+1)\pi-2x$, leading to $x\in\left\{\frac{5\pi}{18},\frac{17\pi}{18},\frac{29\pi}{18}\right\}$. So we get:

$$ x \in \left\{\frac{\pi}{6},\frac{5\pi}{18},\frac{17\pi}{18},\frac{29\pi}{18}\right\}+2\pi\mathbb{Z}.$$

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