Find solutions to $\cot(x)+\csc(x)=\sqrt3$ in range $[0,2\pi]$ What is the best way to do the above? Are there any tricks I should be aware of.
I know how to simplify it to $\dfrac{\cos(x)}{\sin(x)} + \dfrac{1}{\sin(x)} = \sqrt{3}$
so multiplying both sides by $\sin(x)$, we get $\cos(x)+1=\sqrt{3}\sin(x)$.
But I'm stuck from there.
 A: Let's exploit the half-angle identity  
$$\csc x+\cot x=\cot (x/2)$$
Then, we have $\cot(x/2) =\sqrt{3}$ or $\tan(x/2)=\sqrt{3}/3$.
Solutions are
$$x=2\arctan(\sqrt{3}/3)+2n\pi =\pi/3+2n\pi$$
A: Hint:  $\cos(a) \sin(x) - \sin(a) \cos(x) = \sin(x-a)$.  You want $\sin(a)/\cos(a) = \ldots$
A: Just some supplement of user17762's ans:
For $x\neq\pi$,
$$\sqrt3\sin{x}-\cos{x}=1$$
$$\frac{\sqrt{3}}2\sin{x}-\frac12\cos{x}=\frac12$$
$$\sin{x}\cos{\frac{\pi}6}-\sin{\frac{\pi}6}\cos{x}=\frac12$$
$$\sin{x}\cos{\frac{-\pi}6}+\sin{\frac{-\pi}6}\cos{x}=\frac12$$
Using Compound angle formula, we get: 
$$\sin{(x-\frac{\pi}6)}=\frac12=\sin{\frac{\pi}6}$$
$$x-\frac{\pi}6=\frac{\pi}6+2n\pi$$
$$x=\frac{\pi}3+2n\pi$$
where $$n\in\Bbb{Z}$$
A: $$\csc x+\cot x=\sqrt3\iff\csc x-\cot x=\dfrac1{\sqrt3}$$
Adding we get $2\csc x=\dfrac4{\sqrt3}\iff\sin x=\dfrac{\sqrt3}2=\sin\dfrac\pi3>0$
and $\cot x=\sqrt3-\csc x=\cdots=\dfrac1{\sqrt3}\iff\tan x=\sqrt3=\tan\dfrac\pi3>0$
$\implies x$ lies in the first quadrant $\implies x=2n\pi+\dfrac\pi3$ where $n$ is any integer
A: Let $x = 2t$, then we can start at where you left off: 
$\cos x + 1 = \sqrt{3}\sin x\Rightarrow \cos (2t)+1 = \sqrt{3}\sin (2t) \Rightarrow 2\cos^2 t = 2\sqrt{3}\sin t\cos t \Rightarrow 2\cos t(\cos t - \sqrt{3}\sin t) = 0\Rightarrow \cos t = 0 \Rightarrow t = (2n+1)\dfrac{\pi}{2} \Rightarrow x = 2t = (2n+1)\pi, n \in \mathbb{Z}$ or $\cos t - \sqrt{3}\sin t = 0 \Rightarrow 2\cos (t+\frac{\pi}{3})=0 \Rightarrow t+\dfrac{\pi}{3} = (2n+1)\dfrac{\pi}{2} \Rightarrow t = n\pi + \dfrac{\pi}{6} \Rightarrow x = 2t = 2n\pi + \dfrac{\pi}{3}, n \in \mathbb{Z}$. Let $n = 0$ we obtain $x = \pi, \dfrac{\pi}{3}$ are the only $2$ solutions that belong to the range $[0,2\pi]$, and they are the only solutions.
A: Rearranging, we obtain
$$\sqrt3 \sin(x) - \cos(x) = 1 \implies \dfrac{\sqrt3}2 \sin(x) - \dfrac12 \cos(x) = \dfrac12 \implies \sin\left(x-\dfrac{\pi}6\right) = \sin\left(\dfrac{\pi}6\right)$$
This gives us that
$$x-\dfrac{\pi}6 = n\pi + (-1)^n \dfrac{\pi}6 \implies x = n\pi +(-1)^n \dfrac{\pi}6 + \dfrac{\pi}6$$
Further, $\sin(x)\neq 0 \implies x =2n\pi + \pi/3$.
