Помогите решить sin (8*x-pi/6) = sinx

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.

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