Question

Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0

Answer 1

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)