<s>[INST] Solve: $5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$ Solve to find the general value of $x$:
$$5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$$
My Attempt:
$$5(1-\sin^2 (x))-4\sin (x)\cos (x)+3\sin^2 (x)=2$$
$$5-5\sin^2 (x)-4\sin (x)\cos (x)+3\sin^2 (x)=2$$
$$2\sin^2 (x)+4\sin (x)\cos (x)=3$$
 [/INST] 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 $n \in \mathbb{Z}$
$$x=\arctan 1 +n\pi=\frac{\pi(1+4n)}{4}$$
$$\land$$
$$x=\arctan 3 +n\pi$$
</s>