Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$ I'm trying to solve this problem.

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

I have simplified this problem to $$f(x)= 8\cos^2x+6\sin x\cos x+8$$ and tried working with $g(x)= 8\cos^2x+6\sin x\cos x$.
I factored out the $2\cos x$ and rewrote the other factor as a linear combination of cosine.
It reduces down to $$g(x)=10\cos x\cos\left(x-\tan^{-1}\frac{3}{4}\right)$$
But then I'm stuck here. Please help me. Perhaps there's a different way to approach this?

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 $7$
