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

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.
