To solve the equation sin^x + sin^3x = 2sin^2x, we can rewrite sin^3x as (sin^2x)(sinx):

sin^x + (sin^2x)(sinx) = 2sin^2x

Now, let's simplify the equation:

sin^x + sin^2x*sinx = 2sin^2x

Since sin^x is the same as sinx, we can substitute sinx for sin^x:

sinx + sin^2x*sinx = 2sin^2x

Now, let's factor out sinx from the second term on the left side:

sinx(1 + sin^2x) = 2sin^2x

Using the identity sin^2x + cos^2x = 1, we can substitute 1 - cos^2x for sin^2x:

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

Simplifying further:

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

Expanding the left side:

2sinx - sinx*cos^2x = 2 - 2cos^2x

Now, let's move all the terms to one side of the equation:

2sinx - sinx*cos^2x - 2 + 2cos^2x = 0

Combining like terms:

2sinx + cos^2x(2 - sinx) - 2 = 0

Now, we have a quadratic equation in terms of cosx. Let's substitute y = cosx:

2sinx + y^2(2 - sinx) - 2 = 0

Expanding the equation:

2sinx + 2y^2 - y^2*sinx - 2 = 0

Rearranging the terms:

2sinx - y^2*sinx + 2y^2 - 2 = 0

Factoring out sinx:

sinx(2 - y^2) + 2y^2 - 2 = 0

Now, we have a quadratic equation in terms of sinx. Let's substitute z = sinx:

z(2 - y^2) + 2y^2 - 2 = 0

Expanding the equation:

2z - y^2*z + 2y^2 - 2 = 0

Rearranging the terms:

2z - y^2*z + 2y^2 - 2 = 0

Now, we have a quadratic equation in terms of z. We can solve this equation to find the values of z (sinx) that satisfy the original equation.