Question

2cos4x +cos2x=1 решите тригонометрическое уравнение

Answer 1

$2cos4x+cos2x=1$

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

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

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

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

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

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

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

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