Автор - nastyusha163

Помогите пожалуйста решить:

Ответ

Автор - nKrynka
sin(x + π/3)*cos(x - π/3) = 1/4
1/2[(sin(x + π/3 + x - π/3) + sin(x + π/3 - x + π/3)] = 1/4
1/2[sin2x + sin2π/3)] = 1/4
sin2x + √3/2 = 1/2
sin2x = 1/2 - √3/2
sin2x = (1 - √3)/2
2x = (-1)^n*arcsin((1 - √3)/2) + 2πk, k∈Z 
x = [(-1)^n*arcsin((1 - √3)/2)]/2 + πk, k∈Z 













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