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find the absolute minimum and absolute maximum for the given funtion f(x)=x-2sinx between 0 and 2(pi)

Trigonometry
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PLS GIVE A SIMFILICATION TO SOLVE THE PROBLEM
Use the derivative:\[f'(x)=1-2\cos x\]Solve the equation:\[f'(x)=0 \Leftrightarrow 1-2\cos x = 0 \Leftrightarrow \cos x = \frac{ 1 }{ 2 }\]There are two solutions in [0, 2pi]. These are the x-values where f has a (local) extreme. You can calculate the extremes by substituting the solutions of f' in f. Also calculate f(0) and f(2pi) to get the extremes in the endpoints. If you have all the extremes, you can decide what the absolute maximum and minimum values are.

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