Решите неравенство 2cos(3x-pi/3) < корень из 3
cos(3x–pi/3) < sqrt(3)/2 (π/6)+2πk < 3x-(π/3) < (11π/6)+2πk, k∈Z. (π/6)+(π/3)+2πk < 3x < (11π/6)+(π/3)+2πk, k∈Z. (3π/6)+2πk < 3x < (13π/6)+2πk, k∈Z. (π/6)+(2π/3)k < x < (13π/18)+(2π/3)k, k∈Z.