The equation sinx+xcosx=0 has at least one root in -

Let f(x)=sinx+xcosx Consider g(x)=∫0x(sint+tcost)dt=tsint0x=xsinxg(x)=xsinx which is differentiable in0,π. Now, g(0)=0 and g(π)=0 ∴ using Rolles Theorem ∃ atleast one c∈(0,π) such that g'(c)=0 i.e. ccosc+sinc=0 for atleast one c∈(0,π)