Solve: $5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$

Question

Solve to find the general value of $x$: $$5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$$

My Attempt:

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

Answer 1

Since $\sin^2(x)+\cos^2(x)=1$

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

Let $\cos(x)=a, \sin(x)=b$

$5a^2-4ab+3b^2=2a^2+2b^2$

$3a^2-4ab+b^2=0$

$(3a-b)(a-b)=0$

$3a=b$ or $a=b$, which means $3\cos(x)=\sin(x)$ or $\cos(x)=\sin(x)$

$\frac{\sin(x)}{\cos(x)}=\tan(x)=3$ or $\frac{\sin(x)}{\cos(x)}=\tan(x)=1$

You may go from here.

Answer 2

Guide:

$$2\sin^2(x)+4\sin(x)\cos(x)=3$$

$$2\sin(2x)=2+1-2\sin^2(x)$$

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

$$2\sin(2x)-\cos(2x)=2$$

The trick from here should help in solving the problem.

Answer 3

You can divide by $\cos^2 x \quad $ (Since $x=\pi/2$ does not work)

$$\implies 5-4 \tan x+3 \tan^2 x=2 \sec^2 x$$

yet $\sec^2 x=1+\tan^2 x .....$

Continuing you have a quadratic in $\tan x$ which I'm sure you know how to solve...

$$\implies z^2-4z+3=(z-3)(z-1)=0 \quad |\quad z=\tan x $$

So for any $n \in \mathbb{Z}$

$$x=\arctan 1 +n\pi=\frac{\pi(1+4n)}{4}$$ $$\land$$ $$x=\arctan 3 +n\pi$$