sin3x = √3/2; x∈[-3π/2; π].
3x = (-1)ⁿπ/3 + πn, n∈Z
x = (-1)ⁿπ/9 + πn/3, n∈Z
Промежутку [-3π/2; π] принадлежат корни: π/9; 2π/9; 7π/9; 8π/9; -4π/9; -5π/9; -10π/9; -11π/9.
Можно складывать неравенства одного знака:
tg(arccos3/5)=sin(arccos3/5)/cos(arccos3/5)=(4/5)/(3/5)=4/3
cos(arccos3/5)=3/5
cos^2(arccos3/5)=9/25
sin^2(arccos3/5)=16/25
sin(arccos3/5)=4/5