Matematické Fórum

stuart clark · 23. 02. 2017 14:03

Total number of real solution of $\sqrt{2}|\sqrt{\sin^4 x+9\cos^2 x}-\sqrt{\cos^4 x+9\sin^2 x} |= 1$ in $x\in [0,2\pi]$

Bati · 14. 03. 2017 01:31

Hi ↑ stuart clark:,
let $f(x):=\sqrt{\sin^4x+9\cos^2x}-\sqrt{\cos^4x+9\sin^2x}$ , $g:=|f|$ . It is easy to check that $g(x+\tfrac{\pi}2)=g(x)$ , hence it is enough to consider $x\in[0,\tfrac{\pi}2]$ . Now $g(0)=g(\tfrac{\pi}2)=2$ and $g(\tfrac{\pi}4)=0$ , so, by continuity, there are at least 8 solutions in total. In fact, there are no more, because $g$ is monotone on the intervals $(0,\tfrac{\pi}4)$ , $(\tfrac{\pi}4,\tfrac{\pi}2)$ and so on. Indeed, since $\sin x<\cos x$ on $(0,\tfrac{\pi}4)$ and conversely on the next interval, we deduce that $\text{sgn}\, f(x)$ is 1 on $(0,\tfrac{\pi}4)$ and -1 on $(\tfrac{\pi}4,\tfrac{\pi}2)$ . Using that and
$\text{sgn}\,g'(x)=\text{sgn}\,$(\text{sgn}\, f(x))(\sin{2x})\(\frac{2\sin^2x-9}{2\sqrt{\ldots}}+\frac{2\cos^2x-9}{2\sqrt{\ldots}}$\)\nl =-\text{sgn}\,f(x)$ , $x\in(0,\tfrac{\pi}2)$ , the claim is proved.

stuart clark · 27. 03. 2017 12:59

Thanks ↑ Bati:

Matematické Fórum

#1 23. 02. 2017 14:03

Trigonometric equation

#2 14. 03. 2017 01:31

Re: Trigonometric equation

#3 27. 03. 2017 12:59

Re: Trigonometric equation

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