F'(x)=xsin(x). F(x)=∫(xsin(x)dx).

Let u=x and dv=sin(x)dx, du=dx, v=-cos(x)

F(x)=-xcos(x)+∫cos(x)dx=sin(x)-xcos(x)+c, which is what you thought, except for the constant of integration—which is important if we need the zeroes of F(x).

F(0)=c. The zeroes of F(x) would be x=0 (only one solution), but ONLY if c=0.

We have no way of knowing c from the information given in the question, so let’s assume you can find c, then Newton’s Method is one way to solve for c≠0.

We use the iterative formula:

xᵣ₊₁=xᵣ-F(xᵣ)/F'(xᵣ)=xᵣ-(sin(xᵣ)-xcos(xᵣ)+c)/(xsin(xᵣ)).

Starting with r=0 and x₀=3/2 (arbitrary value close to an estimated zero, which can be established graphically). x=3/2 is close to a zero as c gets larger. There are multiple zeroes, and Newton’s Method will only find the nearest zero to x₀.