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
 A: 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
A: 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
$$\sin ^{2} x+2 \cos ^{2} x+3 \sin x \cos x=0 \div \sin ^{2} x \to \\ 1+2 \cot^2 x+3\cot x=0\\ and \div \cos^2x \to \tan^2x+2+3\tan x=0 $$
A: Well, in a case very unusual for these type of questions the equation can be factored.
$\sin^2 x + 2\cos^2x + 3\sin x \cos x = (\sin x + \cos x)(\sin x + 2\cos x)=0$.
so
we have solutions when $\sin x = -\cos x$ or $\sin x=-2\cos x$.  Looking at the unit circle it is clear that these can only occur in the 2nd and 4th quadrants where $\sin$ and $\cos$ are opposite signs.  Furthermore in the quadrants the absolute values of one trig function is increasing from $0$ to $1$ and the other is decreasing from $1$ to $0$ so there will be exactly one solution to each equation in each quadrant.
so there are four solutions:
....
We could do what you did.  In fact that would be my preferred way of doing it.
!!BUT!!!

*

*When you squared both sides you risked adding extraneous solutions.


*Declaring a fourth degree polynomial has four roots doesn't take into account that maybe there are multiple roots or that there might not be real roots.


*$\cos x \ne \sqrt{1 - \sin^2 x}$.  $\cos x=\pm \sqrt {1-\sin^2 x}$
And most importantly.
4)If $t = t_1,t_2,t_3,t_4$ are the four solutions then $\sin x = t_i$ do not have one solution only.  If $|t_1| > 1$ then $\sin x=t_i$ has no solution. And if $|t_i| < 1$ then $\sin x = t_1$ has two solutions.
and least importantly


*$-4 - 9 = -13$ not $-5$.

Let's redo your work
$\sin^2 + \cos^2 = 1$ so
$\sin^2 x+ 2cos^2 x + 3\sin\cos x =$
$1 + \cos^2 x + 3\sin \cos x$.
Not sure why you chose $t = \sin x$ rather than $t=\cos x$ but it shouldn't matter.
$1 + (1 - t^2) \pm 3t\sqrt{1-t^2}=0$
$2-t^2 = \pm 3t\sqrt{1-t^2}$  If we square both sides we need to take note that $1-t^2 \ge 0$ or $t^2 \le 1$.  On the other hand we don't need to worry about the extraneous solution that the we lose the sign of the RHS because we don't know the sign of the RHS.
$4 - 4t^2 + t^4 = 9t^2(1-t^2)$
$10t^4 - 13t^2 + 4=0$
$t^2 = \frac {13 \pm {169-160}}{20} = \frac {13\pm 3}20= \frac 12, \frac 45$.
So four solutions to $t$:  $\pm \frac{\sqrt 2}2; \pm \frac {2\sqrt 5}5$.
But for each $\sin x = t$ there are two values that $x$ could be. So $8$ values?
But no....we have to bear in mind that $\sin^2 x + 2\cos^2 x \ge 0$ so $\sin x\cos x \le 0$ so $\sin x $ and $\cos x$ are opposite signs.  (That slipped me by!)
So when we had $2-t^2 = \pm 3t\sqrt{1-t^2}$ we DO know that that the RHS is positive and that we did get extraneous solutions.
Now we have $\sin x =  \pm \frac{\sqrt 2}2; \pm \frac {2\sqrt 5}5$ AND $\cos x$ is the opposite sign.  There is only one solution for each value.
.....
Moral.  Be careful!
