Solving $3\sin^2x - 4\sin x\cos x + 5\cos^2x = 2$

Question

$3\sin^2x - 4\sin x\cos x + 5\cos^2x = 2$

What I would do is divide the entire equation with $\cos^2x$, but the $2$ on the right side is preventing me from doing that.

Stuck. Please help.

Answer 1

The 2 on the RHS does not stop you from dividing by $\cos^2 x$: $$ 3\tan^2x-4\tan x+5=2\sec^2x=2\tan^2x+2 $$ So $$ \tan^2x-4\tan x+3=0 $$ etc.

Answer 2

Hint: Use the substitution $$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ to obtain a polynomial. You will get $${\frac { \left( {t}^{2}+2\,t-1 \right) \left( 3\,{t}^{2}+2\,t-3 \right) }{ \left( {t}^{2}+1 \right) ^{2}}} =0$$

Answer 3

Rewrite in terms of $\cos^2(x)$:

$$3(1-\cos^2(x))-4\sin(x)\cos(x)+5\cos^2(x)-2=0$$

becomes $$ 1 + 2\cos^2(x)-4\cos(x)\sin(x)=0$$

then we can change this to $$ \sin^2(x)+3\cos^2(x)-4\cos(x)\sin(x)=0$$

factorising:

$$(\sin(x)-\cos(x))(\sin(x)-3\cos(x))=0$$

Can you do the last step?

Answer 4

Additionally one can use on the RHS,

$$2 = 2(\sin^2 x + \cos^2 x)$$

and proceed.

Answer 5

Another alternative: Use $\cos^2x + \sin^2 x=1$, $2\sin x \cos x = \sin 2x$ and $2\cos^2x = \cos 2x +1$. $$3\sin^2x - 4\sin x \cos x + 5\cos^2x = 2$$ $$3\sin^2x+3\cos^2x - 2(2\sin x \cos x) + 2\cos^2x = 2$$ $$3 -2\sin 2x + 2\cos^2x=2$$ $$1- 2 \sin 2x +\cos 2x + 1 = 0$$ $$2=2\sin 2x - \cos 2x$$

Now, use $A\sin(2x+x_0) = A\sin 2x \cos x_0 + A\cos 2x \sin x_0$ an compare coefficients $A\cos x_0=2$ and $A\sin x_0 = -1$ to obtain $A = \sqrt{2^2+1}$ and $\tan x_0 = -0.5$ to rewrite the previous equation as

$$2=A\sin(2x+x_0).$$

It should be easy to solve from here.