Question

2cos5xcos3x+sinx=cos8x

Answer 1

It looks like your equation (it's not an identity) is

2 cos(5x) cos(3x) + sin(x) = cos(8x)

Recall that

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

cos(x - y) = cos(x) cos(y) + sin(x) sin(y)

==>   2 cos(x) cos(y) = cos(x + y) + cos(x - y)

so that

2 cos(5x) cos(3x) = cos(8x) + cos(2x)

Then the equation simplifies to

cos(8x) + cos(2x) + sin(x) = cos(8x)

cos(2x) + sin(x) = 0

Also recall that

cos(2x) = 1 - 2 sin²(x)

so the equation is quadratic in sin(x) and can be factorized:

1 - 2 sin²(x) + sin(x) = 0

2 sin²(x) - sin(x) - 1 = 0

(2 sin(x) + 1) (sin(x) - 1) = 0

Solve for x :

2 sin(x) + 1 = 0   or   sin(x) - 1 = 0

sin(x) = -1/2   or   sin(x) = 1

[x = arcsin(-1/2) + 2nπ   or   x = π - arcsin(-1/2) + 2nπ]   or   x = arcsin(1) + 2nπ

(where n is any integer)

x = -π/6 + 2nπ   or   x = -5π/6 + 2nπ   or   x = π/2 + 2nπ