Question

2cos3x=√3 на промежутке от [-пи; пи]

Answer 1

2cos3x = √3 
cos3x = √3 / 2 
3x = arccos(√3 / 2) + 2πk, k є Z 
3x = ±π/6 + 2πk, k є Z 
x = ±π/18 + 2πk, k є Z