Trebuie calculata prin inductie dar nu stiu exact cum
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x1=f(xo)=xo+cosxo
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xn=f(xn-1)=xn-1+cosxn-1
revenim la functia f(x)=x+cosx
f `(x)=1-sinx>0∀x∈(0,π/2)=> f crescatoare