2cos3x=√3 на промежутке от [-пи; пи]

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

(5/8)×x  +  (45/100)×x = 344

(5/8)×x  + (9/20)×x = 344

(100 + 72)/ 160 = 344

(172 / 160)×x = 344

x = (344×160)/172 = 320