Question

Помогите пожалуйста, срочно нужно
9arccoss²2x - 3π arccos2x - 2π² = 0​

Answer 1

$9\, arccos^22x-3\pi \, arccos2x-2\pi ^2=0\\\\t=arccos2x\ \ ,\ \ 0\leq t\leq \pi \ \ ,\ \ \ 9t^2-3\pi \cdot t-2\pi ^2=0\ \ ,\\\\D=9\pi ^2+4\cdot 9\cdot 2\pi ^2=81\pi ^2\ \ ,\ \ \ t_{1,2}=\dfrac{3\pi \pm 9\pi }{18}\ \ ,\\\\\\t_1=\dfrac{2\pi}{3}\in [\ 0\, ;\, \pi \ ]\ ,\ t_2=-\dfrac{\pi }{3}\notin [\ 0\, ;\, \pi \ ]\\\\\\arccos2x=\dfrac{2\pi}{3}\ \ ,\ \ cos(arccos2x)=cos\dfrac{2\pi}{3}\ \ ,\ \ 2x=-\dfrac{1}{2}\ \ ,\ \ x=-\dfrac{1}{4}\\\\\\Otvet:\ x=-\dfrac{1}{4}\ .$