помогите решить sin (8*x-pi/6)=sinx

2sin(7x/2-π/12)cos(6x/2-π/12)=0sin(7x/2-π/12)=07x/2-π/12=πn,n∈z7x/2=π/12+πn,n∈zx=π/42+2πn/7,n∈zsin(9x/2-π/12)=09x/2-π/12=πk,k∈z9x/2=π/12+πk,k∈zx=π/54+2πk/9,k∈z

 sin (8*x-π/6)=sinx ;sin (8*x-π/6)- sinx  =0 ;2sin(7x/2 -π/12)*cos(9x/2- -π/12) =0 ;а) sin(7x/2 -π/12) =0 ;7x/2 -π/12= π*n ;x =π/42+ 2πn/7 ,n∈Z.---б) cos(9x/2 - π/12) =0  ;9x/2  - π/12 =π/2+π*n ;x = 7π/54 +2πn/9,  n∈Z. ответ : π/42+ 2πn/7  , 7π/54 +2πn/9 , n∈Z.