We can write this as x-sin(x)-1.5=0, then let f(x)=x-sin(x)-1.5. The derivative is f'(x)=1-cos(x). With these we can use Newton's Iteration Method for solving for x iteratively.
xn+1=xn-f(xn)/f'(xn). Let x0=π/2, then:
x1=2.5, x2=2.277..., x3=2.2671..., x4=2.267172, which is stable. So the solution is x=2.267172 approx. This is less than π.
A graph of f(x) will show that the x-intercept is less than π and will provide a rough estimate (accuracy depends on the scale of the graph). Or simply substitute some values of x and note where f(x) changes sign. f(π)=π-1.5>0 and f(0)=-1.5, so the x-intercept occurs between 0 and π.