Question

cos5x = -- корн 3 / 2

Answer 1

$cos5x=- frac{ sqrt{3} }{2} \ 5x=pm arrcos (- frac{ sqrt{3} }{2} )+2 pi k, k in Z\\ 5x=pm ( pi - arrcos ( frac{ sqrt{3} }{2} ))+2 pi k, k in Z\\ 5x=pm ( pi - frac{ pi }{6} )+2 pi k, k in Z\\ 5x=pm frac{ 5pi }{6} +2 pi k, k in Z\\ x=pm frac{ pi }{6} + frac{2}{5} pi k, k in Z$