Saturday, 8 April 2017

trigonometry - How prove this equation has only one solution $cos{(2x)}+cos{x}cdotcos{(sqrt{(pi-3x)(pi+x)}})=0$


Let $x\in (0,\dfrac{\pi}{3}]$.
Show that this equation
$$\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0$$
has a unique solution $x=\dfrac{\pi}{3}$




I try to the constructor $$f(x)=\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0\, , \quad\quad f(\dfrac{\pi}{3})=0$$but I use found this function is not a monotonic function

see wolframpha



Now the key How to prove this function $f(x)$ in $(0,\frac{\pi}{3})$ has no solution, since
$$f\left(\frac{\pi}{6}\right)=\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cos{\left(\dfrac{1}{2}\sqrt{\dfrac{7}{3}}\pi\right)}=-0.138\cdots<0$$
in other words, how to prove that
$$f(x)<0,\forall x\in(0,\dfrac{\pi}{3}) \, .$$

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...