Question

arccos (-1) -arccos 1/2+3 arccos(-√3/2) Умоляю!

Answer 1

arccos (-1) -arccos 1/2+3 arccos(-√3/2)=$\pi -\frac{\pi }{3}+3*\frac{5\pi }{6}=\pi-\frac{\pi }{3}+\frac{5\pi }{2}=\frac{19\pi }{6}$ или 570°