To prove the trigonometric identity sinx + sin2x = 2cos^2x + cosx, we will use trigonometric identities to simplify the left side of the equation and then show that it is equal to the right side.

Starting with the left side of the equation:

sinx + sin2x

Using the double angle identity sin2x = 2sinxcosx:

sinx + 2sinxcosx

Now, we will use the Pythagorean identity sin^2x + cos^2x = 1 to replace sin^2x with 1 - cos^2x:

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

Expanding the expression:

sinx + 2sinxcosx - 2sinxcos^3x

Now, we will simplify the expression further by factoring out a sinx:

sinx(1 + 2cosx - 2cos^2x)

Now, we will use the Pythagorean identity sinx = √(1 - cos^2x) to replace sinx:

√(1 - cos^2x)(1 + 2cosx - 2cos^2x)

Expanding and simplifying the expression gives:

√(1 - cos^2x + 2cosx - 2cos^2x + 2cosx - 4cos^3x)

Using the Pythagorean identity again, cos^2x = 1 - sin^2x, we get:

√(3(1 - sin^2x) - 4(1 - sin^2x) + 2√(1 - sin^2x) - 1)

√(3 - 3sin^2x - 4 + 4sin^2x + 2√(1 - sin^2x) - 1)

√(1 + sin^2x + 2√(1 - sin^2x) - 1)

√(sin^2x + 2√(1 - sin^2x))

Therefore, sinx + sin2x is not equal to 2cos^2x + cosx. In fact, the two expressions do not have an equivalent relationship.