[Math] Find the maximum and minimum value of $\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$

Question

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

Answer 1

The expression can be simplified to $$\cos^2 x + \sin^2 x +2\sin^2 x +2 -3\sin 2x $$ $$=3 -3\sin 2x + 2\frac {1-\cos 2x}{2} $$ $$=4 - 3\sin 2x - \cos 2x $$

Now we know that $$-\sqrt{a^2 +b^2} \leq a\sin \alpha + b \cos \alpha \leq \sqrt {a^2 +b^2} $$

Hope you can take it from here.