Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$

Rewrite as following

$$\begin{align} f(x) & = 8 + 6 \sin(x ) \cos(x ) + 8\cos^2(x) - 4 + 4 \\ &= 12 + 3 \sin(2x) + 4 \cos(2x) \\ &= 12 + 5 \left(\frac{3}{5} \sin(2x) + \frac{4}{5} \cos(2x)\right) \\ &= 12 + 5 \sin(2x + \arctan{\tfrac{4}{3}}) \end{align}$$

Now its easy since $\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \pm 5$.

So maximum value is $17$ and minimum is $7$

write your function as $$f(x)=(\sin(x)+3\cos(x))^2+7$$