1) 2sin x - sin x=0
sin x = 0
x = πn, n∈Z
2) cos 2x cos 4x - sin 2x cos 4x = 1
cos (2x+4x) = 1
6x = πn
x = πn/6, n∈Z
3) Если в таком виде, как написано, то:
sinx - π/9 = 0 или cos3x + 6π/5 = 0
sinx = π/9 cos3x = -6π/5
x = (-1)^n*arcsin(π/9) + πn нет корней, т.к. 6π/5>1
А если (sin (x-π/9))(cos (3x + 6π/5))=0
sin (x-π/9) = 0 или cos (3x + 6π/5) = 0 ⇒ cos (3x + π + π/5) = 0
x-π/9 = πn - cos (3x + π/5) = 0
x = π/9 + πn 3x+π/5 = π/2 + πk
3x = π/2 - π/5 + πk
3x = 3π/10 + πk
x = π/10 + (πk)/3