Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$
My attempt :
\begin{align*}
f(x)&=\dfrac{5}{9\cos^2x-6\sin x\cos x+\sin^2x-6\cos^2x}\\
&= \dfrac{5}{(3\cos x+\sin x)^2-6\cos^2x}
\end{align*}
The problem is if I'm going to use
$$-1\leqslant\sin x\leqslant1\;\text{and}-1\leqslant\cos x\leqslant1$$
I think I need to have only one term.
Edit : I have made some more progress
$$-3\leqslant 3\cos x\leqslant 3$$
$$\therefore -4\leqslant 3\cos x+\sin x\leqslant 4$$
$$ 0\leqslant (3\cos x+\sin x)^2\leqslant 16$$

this $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ can be simplified more neatly, if you use:
$\sin^2(\theta)+ \cos^2(\theta)=1$
$\cos(2\theta)=2\cos^2(\theta)-1$
$\sin(2\theta)=2\sin(\theta)\cos(\theta)$
And the simplified denominator would be in the form:
$$ a\cos(2x)+b\sin(2x)+c$$
Then you can use $$-\sqrt{a^2+b^2} \le a\cos(x) \pm b\sin(x) \le \sqrt{a^2+b^2}$$
