Excuse me: judge the number of zeros of function f (x) = xsinx-3 / 2 in (0, π) and prove it thank you.

f(x)=xsinx-3/2
f'(x)=sinx+xcosx
Let f '(x) = 0 get TaNx = - x, the solution is x0, x0 ∈ (π / 2, π)
(0, x0), f (x) increases, (x0, π), f (x) decreases
f(x)max=f(x0)>f(π/2)>0
f(0)=-3/2

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