So f'(x)=2x+sinx.

F'(x) is odd function, so it is enough to discuss its comparison with 0 on (0, π/2).

On (0, π/2], f' (x) >: 0, so f(x) increases monotonically on (0, π/2].

On [-π/2,0] f' (x)

And f(0)=0- 1=- 1,? f(-π/2)=(-π/2)^2+ 1>； 0,f(π/2)=(π/2)^2- 1>； 0,

So the general graph is as shown in figure (1)? show

G(x)=|f(x)|- 1/2 zero. As long as the image of image |f(x)| has several intersections with the image of y= 1/2.

The image of |f(x)| just folds up the negative part of the original image of f(x). As shown in Figure (2), it can be seen that the number of zeros is four.