Finding the number of solutions to $\sin^2x+2\cos^2x+3\sin x\cos x=0$ with $0\leq x<2\pi$ 
For $0 \leq x<2 \pi$, find the number of solutions of the equation
$$
\sin^2 x+2 \cos^2 x+3 \sin x \cos x=0
$$

I have dealed the problem like this
$\sin ^{2} x+\cos ^{2} x+\cos ^{2} x+3 \sin x \cos x=0$
LET, $\sin x=t ;\quad \sin ^{2} x+\cos ^{2} x=1$
$t^{2}+2-2 t^{2}+3 t \sqrt{1-t^{2}}=0$
$\left(t^{2}+2\right)^{2}=9 t^{2}\left(1-t^{2}\right)$
$t^{4}+4 t^{2}+h=9 t^{2}-9 t^{4}$
$10 t^{4}-5 t^{2}+4=0$
So the number of solution must be 4
P.s- Any other approach will be greatly appreciated!
correct me if I am wrong

Other idea to solve $a\sin^2 x+b \cos^2x+ c \sin x \cos x =d $ is divide by $\cos^2x$ or $\sin^2x$ to turn to quadratic like equation of $\tan x$ function
$$\sin ^{2} x+2 \cos ^{2} x+3 \sin x \cos x=0 \div \sin ^{2} x \to \\ 1+2 \cot^2 x+3\cot x=0\\ and \div \cos^2x \to \tan^2x+2+3\tan x=0 $$
