ответ: {x= (arccos1/3)/4 + kπ/2, k ∈ Z
{x= - (arccos1/3)/4 + kπ/2, k ∈ Z
{x= π/6 + kπ/2, k ∈ Z
{x= π/3 + kπ/2, k ∈ Z
пусть cos4x = t, тогда
6t² + t - 1=0
D= 1 + 24 = 25
t1 = 1/3
t2 = - 1/2
cos4x = 1/3 cos4x = -1/2
{x= (arccos1/3)/4 + kπ/2, k ∈ Z {x= π/6 + kπ/2, k ∈ Z
{x= - (arccos1/3)/4 + kπ/2, k ∈ Z {x= π/3 + kπ/2, k ∈ Z
