Найдите корни уравнения tg4x=√3/3 на отрезке [-pi ; pi /2]

Tg4x=√3/3  4x=π/3+πk  x=π/12+πk/4
надо найти решения ∈[ -π;π]
k=0   π/12
k=1  π/12+π/4=π/3
k=-1  π/12-π/4=-π/6
k=2   π/12+π/2=7/12π
k=-2  π/12-π/2=-5π/12
k=3   π/12+3/4π=5π/6
k=-3  π/12-3π/4=-π/3
k=4  π/12+4π/4>π  
k=-4  π/12-π=-11π/12
k=-5   π/12-5π/4=-14π/12<-π

решения выделены жирным.