Discuss the extremum of `f(x)=sinx(1+cosx),x in (0,pi/2)`

Explanation

Let f(x) =sin x sin x(1+cosx)
`therefore` f(x)=cos 2x+cos x
and f(x) =-2sin 2x-sin x =-(2sin 2x+sinx)
For maximum or minimum value of f(x),f(x)=0 i.e
or cos 2x+cosx=0
or cos x=-cos 2x
or cos x= cos `(pi pm 2x)`
`x=pi pm 2x` or `x =(pi)/(3)`
Now `f(pi)/(3)=-2sin (2pi)/(3)-sin (pi)/(3)`
`=-2sqrt(3)/(2)-sqrt(3)/(2)=-3sqrt(3)/(2)`=-Ve
Hence f(x) has maxima at `x=(pi)/(3)`