Решите уравнение, ) 1) sin5x-sin6x 2) sin3x-√3cos2x-sinx=0 заранее

1.
sin5x = sin6x
sin5x-sin6x=0
-2cos(11x/2)•sin(x/2)=0
1) cos(11x/2)=0   =>  11x/2=π/2+πn  =>  x= π/11+(2π/11)•n, n∊Z
2) sin(x/2)=0  =>  x/2=πn  =>  x=2πn, n∊Z.
2.
sin3x-sinx=2
sinxcos2xcos2x(2sinx-sqrt(3))=0
cos2x=0 
2x=П/2*(2n+1)  
x=П*((2n+1)/4)
sinx=sqrt(3)/2
x=П/3+2Пn, n∊Z.
x=2/3П+2Пn, n∊Z.