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
$$-3\leqslant 3\cos x\leqslant 3$$
$$\therefore -4\leqslant 3\cos x+\sin x\leqslant 4$$
$$ 0\leqslant (3\cos x+\sin x)^2\leqslant 16$$
 A: The denominator can be written
$$\sin^2x-6\sin x\cos x+3\cos^2x
=\frac{1-\cos 2x}2-3\sin 2x+3\frac{\cos 2x+1}2
\\=2+\cos2x-3\sin2x,$$
which varies continuously in $[2-\sqrt{10},2+\sqrt{10}]$.
Hence as the interval straddles $0$, the range of the function is
$$\left(-\infty,\frac5{2-\sqrt{10}}\right]\cup\left[\frac5{2+\sqrt{10}},\infty\right)$$
A: Another way:
$$y(\sin^2x -6\sin x\cos x+3\cos^2x)=5$$
Divide both sides by $\cos^2x$
$$y\tan^2x-6y\tan x+3y=5(1+\tan^2x)$$
Rearrange to form a quadratic equation in $\tan x$ which is real
So, the discriminant must be $\ge0$
A: $$f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2 x}=\frac{10}{2\sin^2 x-12 \sin x \cos x+6 \cos^2 x}.$$
$$f(x)=\frac{10}{1-\cos 2x-6 \sin 2x +3(1+\cos 2x)}=\frac{10}{4+2\cos 2x-6 \sin 2x}$$
$$\implies f(x)=\frac{5}{2+\cos 2x-3 \sin 2x}=\frac{5}{2+\sqrt{10}(\cos 2x-3\sin 2x)/\sqrt{10}}$$
$$\implies f(x)=\frac{5}{2+\sqrt{10}\cos(2x+a)}~~~~\color{red}{(1)}$$
$$\implies f_{\min}=\frac{5}{2+\sqrt{10}},\quad f_{\max}=\frac{5}{2-\sqrt{10}}.$$
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 $(-\infty,\frac{5}{2-\sqrt{10}}] \cup [\frac{5}{2+\sqrt{10}}, \infty )$.
A: 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 $$-\sqrt{a^2+b^2} \le a\cos(x) \pm b\sin(x) \le \sqrt{a^2+b^2}$$
