f(x) = sin2x - √3x.
f '(x) = 2cos2x - √3.
f '(x) = 0, т.е. 2cos2x - √3 = 0,
2cos2x = √3,
cos2x = √3/2,
2x = +-π/6 + 2πk, k ∈ Z,
x = +-π/12 + πk, k∈Z.
ответ: x = +-π/12 + πk, k∈Z.
f(x) = sin2x - √3x.
f '(x) = 2cos2x - √3.
f '(x) = 0, т.е. 2cos2x - √3 = 0,
2cos2x = √3,
cos2x = √3/2,
2x = +-π/6 + 2πk, k ∈ Z,
x = +-π/12 + πk, k∈Z.
ответ: x = +-π/12 + πk, k∈Z.