Question

2cos4x +cos2x=1 решите тригонометрическое уравнение

Answer 1

$2cos4x+cos2x=1$

$2(2cos^22x-1)+cos2x=1$

$4cos^22x+cos2x-3=0$

$cos2x=t,t \in [-1;1]$

$4t^2+t-3=0$

$t_1=-1;t_2=\frac{3}{4}$

$cos2x=-1;cos2x=\frac{3}{4}$

$2x=\pi+2\pi n,2x=+-arccos\frac{3}{4}+2\pi k,k,n \in Z$

$x=\frac{\pi}{2}+\pi n,x=+-\frac{1}{2}arccos\frac{3}{4}+\pi k,k,n \in Z$