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

Question

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

Answer 1

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

Answer 2

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

$$\sin ^{2} x+2 \cos ^{2} x+3 \sin x \cos x=0 \div \sin ^{2} x \to \\ 1+2 \cot^2 x+3\cot x=0\\ and \div \cos^2x \to \tan^2x+2+3\tan x=0 $$

Answer 3

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!!!

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

2. 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.

3. $\cos x \ne \sqrt{1 - \sin^2 x}$. $\cos x=\pm \sqrt {1-\sin^2 x}$

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

5. $-4 - 9 = -13$ 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!