The x intercepts are when ln(1+x)=sin²x. One obvious solution is x=0. There are two other zeroes in the given range. The graph of the function shows that x=1 and x=1.7 are close to the zeroes. Newton’s Method can be used to find these. First we need to differentiate the function: f'(x)=1/(1+x)-2sin(x)cos(x)=1/(1+x)-sin(2x). The iterative equation is x=x-(ln(x)-sin²x)/(1/(1+x)-sin(2x)). If we put x=1 into the right-hand side we get x=1-(0-sin²(1))/(1/2-sin(2))=0.9635.... We then resubmit x with this value and repeat the process until we get a consistent result. We get 0.9643632012 to 10 decimal places.

Now we start with x=1.7. We get x=1.683885012 to 9 decimal places.