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

• $-\sqrt{a^2+b^2} \le a\cos(x) \pm b\sin(x) \le \sqrt{a^2+b^2}$ – UmbQbify Jul 13 '20 at 9:38

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

$$\left(-\infty,\frac5{2-\sqrt{10}}\right]\cup\left[\frac5{2+\sqrt{10}},\infty\right)$$

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 $$\ge0$$

• There is still significant work between this hint and the solution. – Yves Daoust Jul 13 '20 at 10:13
• @YvesDaoust, Thta's deliberate. But not more than a little algebraic manipulation like math.stackexchange.com/questions/174905/…. Is it the reason for downvoting ? – lab bhattacharjee Jul 13 '20 at 10:16
• Yes, I don't think that this is enough for the OP. – Yves Daoust Jul 13 '20 at 10:17

$$f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2 x}=\frac{10}{2\sin^2 x-12 \sin x \cos x+6 \cos^2 x}.$$ $$f(x)=\frac{10}{1-\cos 2x-6 \sin 2x +3(1+\cos 2x)}=\frac{10}{4+2\cos 2x-6 \sin 2x}$$ $$\implies f(x)=\frac{5}{2+\cos 2x-3 \sin 2x}=\frac{5}{2+\sqrt{10}(\cos 2x-3\sin 2x)/\sqrt{10}}$$ $$\implies f(x)=\frac{5}{2+\sqrt{10}\cos(2x+a)}~~~~\color{red}{(1)}$$ $$\implies f_{\min}=\frac{5}{2+\sqrt{10}},\quad f_{\max}=\frac{5}{2-\sqrt{10}}.$$ 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 $$(-\infty,\frac{5}{2-\sqrt{10}}] \cup [\frac{5}{2+\sqrt{10}}, \infty )$$.

• $$2\cos2x-3\sin2x\le\sqrt{2^2+3^2}\cos(2x+\arccos\dfrac3{\sqrt{13}})$$ right? – lab bhattacharjee Jul 13 '20 at 10:02
• What about the negative values ?? – Yves Daoust Jul 13 '20 at 10:07
• Even worse now ! – Yves Daoust Jul 13 '20 at 10:14

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 $$-\sqrt{a^2+b^2} \le a\cos(x) \pm b\sin(x) \le \sqrt{a^2+b^2}$$