To solve this equation, we can simplify the left side first:

Sin5x - 2sinx(cos4x + cos2x) = sinx

Using the double angle formula for cosine, we have:

Sin5x - 2sinx(2cos^2(2x) - 1 + 2cos^2(x) - 1) = sinx

Simplifying further:

Sin5x - 2sinx(2cos^2(2x) + 2cos^2(x) - 2) = sinx

Distributing the -2sinx:

Sin5x - 4sinx(cos^2(2x) + cos^2(x) - 1) = sinx

Expanding the square terms:

Sin5x - 4sinx(2cos^4(x) - 2cos^2(x) + 1 + cos^2(x) - 1) = sinx

Simplifying the square terms:

Sin5x - 4sinx(2cos^4(x) - cos^2(x)) = sinx

Expanding the square terms again:

Sin5x - 8sinx(cos^4(x) - cos^2(x)) = sinx

Factoring out sinx:

sinx(1 - 8(cos^4(x) - cos^2(x))) = sinx

Now we have two possibilities:

1 - 8(cos^4(x) - cos^2(x)) = 1

Simplifying:

8(cos^4(x) - cos^2(x)) = 0

Dividing by 8:

cos^4(x) - cos^2(x) = 0

Factoring out cos^2(x):

cos^2(x)(cos^2(x) - 1) = 0

This gives us two possible solutions:

cos^2(x) = 0 or cos^2(x) - 1 = 0

For cos^2(x) = 0, we have:

cos(x) = 0

x = π/2 + nπ, where n is an integer.

For cos^2(x) - 1 = 0, we have:

cos^2(x) = 1

cos(x) = ±1

x = 2nπ or x = (2n + 1)π, where n is an integer.

Therefore, the solutions to the equation are:

x = π/2 + nπ, x = 2nπ, or x = (2n + 1)π, where n is an integer.