Find the maximum and minimum value of $\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$ Find the maximum and minimum value of $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$.
i simplified and reach to expression as follows :
$5 + 2\sin(x)[\sin(x)-3\cos(x)]$. How do i go from here?
Thanks

HINT:
Let $y=\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$
Divide both sides by $\cos^2x$ and use $\sec^2x=1+\tan^2x$ to form a Quadratic Equation in $\tan x$
Now as $\tan x$ is real, the discriminant must be $\not<0$
Or divide both sides by $\sin^2x$ to form a Quadratic Equation in $\cot x$
