2sin2x-1=0
2sin2x=1
sin2x=1/2
2x=π/6 + 2πn, n ∈ Z 2x = 5π/6 + 2πk, k∈Z
x₁ = π/12 + πn, n ∈ Z x₂ = 5π/12 + πk, k∈Z
ответ: π/12 + πn, n ∈ Z ; 5π/12 + πk, k∈Z
