3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ $-\sqrt{a^2+b^2} \le a\cos(x) \pm b\sin(x) \le \sqrt{a^2+b^2}$ $\endgroup$
    – UmbQbify
    Commented Jul 13, 2020 at 9:38

4 Answers 4

5
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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 $\ge0$

$\endgroup$
3
  • $\begingroup$ There is still significant work between this hint and the solution. $\endgroup$
    – user65203
    Commented Jul 13, 2020 at 10:13
  • $\begingroup$ @YvesDaoust, Thta's deliberate. But not more than a little algebraic manipulation like math.stackexchange.com/questions/174905/…. Is it the reason for downvoting ? $\endgroup$ Commented Jul 13, 2020 at 10:16
  • $\begingroup$ Yes, I don't think that this is enough for the OP. $\endgroup$
    – user65203
    Commented Jul 13, 2020 at 10:17
1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ $$2\cos2x-3\sin2x\le\sqrt{2^2+3^2}\cos(2x+\arccos\dfrac3{\sqrt{13}})$$ right? $\endgroup$ Commented Jul 13, 2020 at 10:02
  • $\begingroup$ What about the negative values ?? $\endgroup$
    – user65203
    Commented Jul 13, 2020 at 10:07
  • $\begingroup$ Even worse now ! $\endgroup$
    – user65203
    Commented Jul 13, 2020 at 10:14
0
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .