Answer :

Condition (1):


Since, f(x) = sinx-sin2x is a trigonometric function and we know every trigonometric function is continuous.


f(x) = sinx-sin2x is continuous on [0,2π].


Condition (2):


Here, f’(x)= cosx-2cos2x which exist in [0,2π].


So, f(x)= sinx-sin2x is differentiable on (0,2π)


Condition (3):


Here, f(0)= sin0-sin0 = 0


And f(2π)=sin(2π)-sin(4π) =0


i.e. f(0)=f(2π)


Conditions of Rolle’s theorem are satisfied.


Hence, there exist at least one cϵ(0,2π) such that f’(c)=0


i.e. cosx-2cos2x =0


i.e. cosx-4cos2x+2=0


i.e. 4cos2x-cosx-2=0


i.e.


i.e. c=32° 32’ or c=126°23’


Value of c=32°32’ϵ(0,2π)


Thus, Rolle’s theorem is satisfied.


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