Question

Помогите решить sin (8*x-pi/6)=sinx

Answer 1

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

Answer 2

 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.
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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.