Question

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

Answer 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;