2sinx= -1 2cos2x= -2 из кореня 2 sinx=1 sin 2x=1 sin x/2 = 1 2sin x/2=1

1) sinx =1/2 ; x= π/6 +π*n ;n∈Z. 2) 2cos2x = -2; cos2x = -1 ; 1+cos2x =0 ; 2cos²x  =0 ; cosx=0; x =π/2+ π*n , n∈Z . 3) √2*sinx =1; sinx =1/√2 ; x = (-1)^(n)*π/4 +π* , n∈Z 4) sin2x =1 ; 2x =π/2+2π*n  ,n∈Z  ; x = π/4 +π*n , n∈Z . 5) sin(x/2) =1; x/2 =π/2+2π*n  , n∈Z ; x  =π+4π*n , n∈Z . 6)2sin(x/2) = 1 ; sin(x/2) =1/2; x/2= π/6 +π*n ;n∈Z; x= π/2 +2π*n ;n∈Z;