Find the difference between the greatest and least values of the function `f(x)=sin2x-x on[-pi/2,pi/2]dot`

Explanation

`f(x)=sin 2 x-x Rightarrow f^{prime}(x)=2 cos 2 x-1`


`therefore f^{prime}(x)=0 Rightarrow 2 cos 2 x-1=0 Rightarrow cos 2 x=frac{1}{2}`


`Rightarrow 2 x=frac{pi}{3},-frac{pi}{3} Rightarrow x=frac{pi}{6},-frac{pi}{6}`



Now `{f}(-frac{pi}{2})=frac{pi}{2}, {f}(frac{pi}{2})=-frac{pi}{2}`



`f(-frac{pi}{6})=-frac{sqrt{3}}{2}+frac{pi}{6}, f(frac{pi}{6})=frac{sqrt{3}}{2}-frac{pi}{6}`



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