Given the function f(x)=2sinx+cos(2x) in the interval `[0,2pi]`

Taking the first derivative we get,

f'(x)=2cosx-2sin(2x)

Inorder to find the critical points we have to equate f'(x)=0.

So we get,

2cosx-2sin(2x)=0

i.e. cosx-2sinxcosx=0

cosx(1-2sinx)=0

implies,...

(The entire section contains 147 words.)

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