Respuesta :
Recall the angle sum identity for cosine:
cos(x + y) = cos(x) cos(y) - sin(x) sin(y)
cos(x - y) = cos(x) cos(y) + sin(x) sin(y)
==> sin(x) sin(y) = 1/2 (cos(x - y) - cos(x + y))
Then rewrite the equation as
sin(4x) sin(5x) + sin(4x) sin(3x) - sin(2x) sin(x) = 0
1/2 (cos(-x) - cos(9x)) + 1/2 (cos(x) - cos(7x)) - 1/2 (cos(x) - cos(3x)) = 0
1/2 (cos(9x) - cos(x)) + 1/2 (cos(7x) - cos(3x)) = 0
sin(5x) sin(-4x) + sin(5x) sin(-2x) = 0
-sin(5x) (sin(4x) + sin(2x)) = 0
sin(5x) (sin(4x) + sin(2x)) = 0
Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
Rewrite the equation again as
sin(5x) (2 sin(2x) cos(2x) + sin(2x)) = 0
sin(5x) sin(2x) (2 cos(2x) + 1) = 0
sin(5x) = 0 or sin(2x) = 0 or 2 cos(2x) + 1 = 0
sin(5x) = 0 or sin(2x) = 0 or cos(2x) = -1/2
sin(5x) = 0 ==> 5x = arcsin(0) + 2nπ or 5x = arcsin(0) + π + 2nπ
… … … … … ==> 5x = 2nπ or 5x = (2n + 1)π
… … … … … ==> x = 2nπ/5 or x = (2n + 1)π/5
sin(2x) = 0 ==> 2x = arcsin(0) + 2nπ or 2x = arcsin(0) + π + 2nπ
… … … … … ==> 2x = 2nπ or 2x = (2n + 1)π
… … … … … ==> x = nπ or x = (2n + 1)π/2
cos(2x) = -1/2 ==> 2x = arccos(-1/2) + 2nπ or 2x = -arccos(-1/2) + 2nπ
… … … … … … ==> 2x = 2π/3 + 2nπ or 2x = -2π/3 + 2nπ
… … … … … … ==> x = π/3 + nπ or x = -π/3 + nπ
(where n is any integer)