Question

Помогите решить 6 номер! Срочно нужно.

Answer 1

√3sin2x=cos2x
√3sin2x-cos2x=0/cos2x
√3tg2x-1=0
√3tg2x=1
tg2x=1/√3
2x=π/6+πk,k∈z
x=π/12+πk/2,k∈z
x=π/12
x=π/12+π/2=7π/12
x=π/12+π=13π/12