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, cosx=0 and 1-2sinx=0

So cosx=0 implies x=pi/2, 3pi/2 in the interval [0,2pi].

1-2sinx=0 implies, x=pi/6, 5pi/6.

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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, cosx=0 and 1-2sinx=0

So cosx=0 implies x=pi/2, 3pi/2 in the interval [0,2pi].

1-2sinx=0 implies, x=pi/6, 5pi/6.

Therefore the critical points are: `pi/2, (3pi)/2,pi/6,(5pi)/6`

Now taking the second derivative of f(x) we get,

f''(x)=-2sinx-4cos(2x)

Applying second derivative test we have,

f''(pi/2)=4>0

f''(3pi/2)=4>0

f''(pi/6)=-3<0

f''(5pi/6)=-3<0

Therefore the function has relative minima at ```x=pi/2, (3pi)/2`

and the minimum value is f(x)=1

The function has relative maxima at `x=pi/6,(5pi)/6`

and the maximum value is f(x)=1.5