Question

Решите уравнение?chj

Answer 1

$6cos^2x-9cosx+3=0 \D=81-4*6*3=9=3^2 \cosx= frac{9+3}{12}=1 \x_1=0+2pi n=2pi n \cosx=frac{9-3}{12}=frac{1}{2} \x_2= frac{pi}{3}+2pi n \x_3= -frac{pi}{3}+2pi n$