Question

cos2x=1-cos[n/2-x] решите уравнение

Answer 1

cos2x =1 - cos(π/2 -x) ;
1 -2sin²x =1 -sinx ;
2sin²x =sinx ;
2sinx(sinx -1/2) =0 ;
[ sinx = 0 ; sinx -1/2 =0. [ x = π*n ;x =(-1)^(n) *π/6 +π*n . n∈Z .