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?
 A: 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:

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

A: Hint: It is equivalent to
$$\frac12\cos x+\frac{\sqrt 3}2 \sin x=2\sin x\cos x.$$
A: Hint: $\tan(\pi/6) = 1/\sqrt{3}$.
Edited to add: It's also equivalent to $1 + \sqrt{3}\tan(x) = 4 \sin(x)$.
A: Simply having a catalog of special trig values can help. I saw the $\sqrt{3}$ here, and recalled that $\frac{\sqrt{3}}{2}$ is a special trig value for cosine: $\cos(\pi/6)$. And fortunately, $\sin(\pi/6)$ is $\frac12$, and you can just see that therefore $\pi/6$ is one solution.
But to look for more, your equation is equivalent to $$\frac{1}{s}+\sqrt{\frac{3}{1-s^2}}=4$$ with $s=\sin(x)$. Subtract the $1/s$ to the other side, square, and clear denominators to get a 4th degree polynomial equation in $s$. But you know one root is $s=\sin(\pi/6)=\frac12$. So then you can factor and be left with a cubic, whose roots will be all that much easier to find.
