solve equation for x

sin^2 x=du/dx-cos^2x*dv/dx

u=2x^3+3x^2

v=x^2

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To solve the equation, we'll have to differentiate u and v, with respect to x.

du/dx = d(2x^3+3x^2)/dx

du/dx = 6x^2 + 6x

dv/dx = d(x^2)/dx

dv/dx = 2x

We'll substitute du/dx and dv/dx by their expression.

(sin x)^2 = 6x^2 + 6x - 2x*(cos x)^2

We'll move 2x*(cos x)^2 to the left side.

(sin x)^2 + 2x*(cos x)^2 = 6x(x+1)

But, (cos x)^2 = 1 - (sin x)^2

(sin x)^2 + 2x - 2x(sin x)^2 = 6x(x+1)

(sin x)^2(1 - 2x) = 2x(3x+3-1)

(sin x)^2(1 - 2x) = 2x(3x+2)

(sin x)^2 = 2x(3x+2)/(1 - 2x)

sin x = 0 for x = 0

2x(3x+2)/(1 - 2x)= 0 for x = 0

**The common solution of the given equation is x = 0.**

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