2cos3x=корень из3 [0;п]
[m]2cos3x=\sqrt{3}[/m] [m]cos3x=\frac{\sqrt{3}}{2}[/m] [m]3x= ± arccos(\frac{\sqrt{3}}{2})+2πn, n ∈ [/m][b]Z[/b] [m]3x= ± \frac{π}{6}+2πn, n ∈ [/m][b]Z[/b] [m]x= ±\frac{π}{18}+\frac{2π}{3}n, n ∈ [/m][b]Z[/b]