$2\sin x=3\cot x$ from $0^\circ$ to $360^\circ$ I used two slightly different approaches to solve this.  First approach gives 2 correct solutions,  second approach gives 4 solutions of which 2 are correct and 2 wrong,  I just cannot figure out why I'm getting 2 extra wrong answers with second approach. 


 A: When you write $$2\sin^2 x=3\sqrt{1-\sin^2 x}$$ $$\left(\frac{2\sin^2 x}{3}\right)^2=1-\sin^2 x$$ you're really saying $$2\sin^2 x=3\sqrt{1-\sin^2 x}\Rightarrow\left(\frac{2\sin^2 x}{3}\right)^2=1-\sin^2 x$$ It's false that $$\left(\frac{2\sin^2 x}{3}\right)^2=1-\sin^2 x\Rightarrow 2\sin^2 x=3\sqrt{1-\sin^2 x}$$ because $f(x)=x^2$ isn't an injective function. From that, you get that all the solutions to the first equation are solutions to the last equation, but not all the solutions to the last equation are solutions to the first equation.
A: Going from your 3rd line
$$ 2\sin^2 x = 3\sqrt{1-\sin^2 x} $$
to your 4th line
$$ \left(\frac{2\sin^2}{3}\right)^2 = 1 - \sin^2 x $$
you have taken the square of both sides of the equation. However, taking squares is dangerous, because you're potentially introducing spurious solutions when you take square roots later -- going from $y^2$ to $\pm y$ you have to test both solutions to see which one(s) work.
As a very simple motivating example, consider
$$x=3$$
You can square both sides to get
$$x^2=9$$
but when you take the square root and write
$$x=\pm3$$
only one possibility satisfies the original equation.
