Find the maximum and minimum value of `f(x)=sinx+1/2cos2x in[0,pi/2]dot`

Explanation

Given `f(x)=sin x+1 / 2 cos 2 x`


`=f^{prime}(x)=cos x-sin 2 x`


Now, `f^{prime}(x)=0` gives `cos x-sin 2 x=0`


`Rightarrow cos x(1-2 sin x)=0`


`Rightarrow cos x=0,(1-2 sin x)=0`


`Rightarrow cos x=0, sin x=1 / 2`


`Rightarrow x=pi / 6, pi / 2`


Now, `f(0)=1 / 2`,


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