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
 A: 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. 
A: $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$
$$-3\sin(2x)+2\dfrac{1-\cos2x}{2}+3$$
$$-3\sin(2x)-\cos(2x)+4$$
now use
$$|a\sin\alpha+b\cos\alpha|\leq\sqrt{a^2+b^2}$$
A: 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 $\cot x$
A: $$
\begin{align}
\frac{d}{dx} \cos^2x-6\sin(x)\cos(x)+3\sin^2x+2 &= -6 \cos(2 x) + 2 \sin(2 x)
\end{align}
$$
as a result of a simple differential calculation; now you'd like the $RHS$, namely $ -6 \cos(2 x) + 2 \sin(2 x)$, to be $0$ as to find the local minima and maxima of the function.
$$
\begin{align}
-6 \cos(2 x) + 2 \sin(2 x) &= 0 \iff \\
x &= \pi n + \arctan \big( \frac{1}{3} (-1 - \sqrt{10} \big) \bigvee \\
x &= \pi n + \arctan \big( \frac{1}{3} (-1 + \sqrt{10} \big) \\
\forall &n \in \mathbb{Z}
\end{align}
$$
From here you can find the absolute minima and maxima by several methods.
