Решить тригонометрическое уравнение: в) .

В) .
2*{[cos(x + 3x)/2] * [cos(x - 3x)/2]}= cos2x
2*[cos2x*cosx] - cos2x = 0
cos2x(2cosx - 1) = 0
1)  cos2x = 0
2x = π/2 + πn, n∈Z
x1 = π/4 + (πn)/2, n ∈Z
2)  2cosx - 1 = 0
cosx = 1/2
x = (+ -) arccos(1/2) + 2πk, k∈Z
x2 = (+ -)*(π/3) + 2πk, k∈Z