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
$$-3\leqslant 3\cos x\leqslant 3$$
$$\therefore -4\leqslant 3\cos x+\sin x\leqslant 4$$
$$ 0\leqslant (3\cos x+\sin x)^2\leqslant 16$$

$$f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2 x}=\frac{10}{2\sin^2 x-12 \sin x \cos x+6 \cos^2 x}.$$
$$f(x)=\frac{10}{1-\cos 2x-6 \sin 2x +3(1+\cos 2x)}=\frac{10}{4+2\cos 2x-6 \sin 2x}$$
$$\implies f(x)=\frac{5}{2+\cos 2x-3 \sin 2x}=\frac{5}{2+\sqrt{10}(\cos 2x-3\sin 2x)/\sqrt{10}}$$
$$\implies f(x)=\frac{5}{2+\sqrt{10}\cos(2x+a)}~~~~\color{red}{(1)}$$
$$\implies f_{\min}=\frac{5}{2+\sqrt{10}},\quad f_{\max}=\frac{5}{2-\sqrt{10}}.$$
when $\cos (\cdots)=\mp 1$.
These are only local maximum.and minimum. But $f(x)$ $\color{red}{(1)}$ can take values close to $\pm \infty$ real value so the Range it's range is $(-\infty,\frac{5}{2-\sqrt{10}}] \cup [\frac{5}{2+\sqrt{10}}, \infty )$.
