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

Here is another way to reach the goal,
The equation reduces to:
$$\begin{array}{l}
\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x+\cos ^{2} x+\sin x \cos x=0 \\
(\sin x+\cos x)^{2}+\cos x(\sin x+\cos x)=0
\end{array}$$
\begin{array}{l}
(\sin x+\cos x)(\sin x+2 \cos x)=0 \\
\Longrightarrow \tan x=-1 \text { and } \tan x=-2
\end{array}
$\tan x$ has a period of $\pi,$ Hence, it takes each value twice in an interval of $2 \pi .$ So the answer is 4
