Question

cos 3x=2 под корнем /2

Answer 1

$cos 3x=\frac{\sqrt{2} }{2}$

$3x=бarccos\frac{\sqrt{2} }{2}+2\pi n,$ $n$ ∈ $Z$

$3x=б\frac{\pi }{4} +2\pi n,$ $n$ ∈$Z$

$x=б\frac{\pi }{12} +\frac{2\pi n}{3} ,$ $n$ ∈ $Z$