Т. к. sin(x) ≡ cos( (π/2) - x), поэтому имеем
cos(3x) + cos( (π/2) - x) = 0,
Воспользуемся формулой "сумма косинусов":
cos(A) + cos(B) ≡ 2*cos( (A+B)/2)*cos( (A-B)/2 ),
cos(3x) + cos( (π/2) - x) ≡ 2*cos( (3x + (π/2) -x)/2 )*cos( (3x - (π/2) + x)/2)≡
≡ 2*cos( x + (π/4))*cos( 2x - (π/4) ) = 0,
1) cos( x + (π/4)) = 0,
или
2) cos( 2x - (π/4) ) = 0.
1) x + (π/4) = (π/2) + π*k, k∈Z,
x = (π/2) - (π/4) + π*k = (π/4) + π*k,
2) 2x - (π/4) = (π/2) + π*n, n∈Z,
2x = (π/2) + (π/4) + π*n = (3π/4) + π*n,
x = (3π/8) + (π*n/2).