Trig Equations Using Identities How would you solve:

$2\csc^2x=3\cot^2x-1$

I said:


*

*Turn the cosecant to $1+\cot^2~x$.


*Distribute to get $3=\cot^2~x$.


*Turn it into tan. To get $\tan x=\frac{1}{\pm \sqrt3}$.

Is this correct?
 A: Your method is ok.
Another approach:
$$2\text{cosec}^2x = 3(\text{cosec}^2x -1) -1 = 3\text{cosec}^2x -4 $$
$$  \text{cosec}^2x-4= 0$$
$$ \frac{1}{\sin^2x} = 4$$
$$ \sin^2x = \frac{1}{4}$$
$$ x = k\pi \pm\frac{\pi}{6} $$
A: Notice, we can also solve this as follows $$2\csc^2 x=3\cot^2 x-1$$ $$\implies \frac{2}{\sin^2 x}=\frac{3\cos^2 x}{\sin^2x}-1$$ $$\implies 2=3\cos^2 x-\sin^2x$$ $$\implies 3\cos^2 x-(1-\cos^2x)=2$$ $$\implies 4\cos^2 x=3$$ $$\implies \cos^2 x=\frac{3}{4}=\left(\frac{\sqrt{3}}{2}\right)^2$$ $$\implies \cos^2 x=\left(\cos\frac{\pi}{6}\right)^2$$ Now, writing the general solution as follows 
$$\implies \color{blue}{x=n\pi\pm\frac{\pi}{6}}$$ Where, $\color{blue}{\text{n is any integer}}$
A: Your approach is correct. 
For convenience and speed in latexing  I abbreviate/algebraize trig functions, pushing their arguments into suffix position:
$$  \frac{2}{s_x^2}= 3(\frac{1}{s_x^2} -1) -1 = \frac{3}{s_x^2} -4 $$
Simplifying you get:
$$ s_x = \pm \frac12  ;\,  x = n\pi \pm\frac{\pi}{6}. $$
