Let y=2x. Since 0≤x≤π, we have 0≤y≤2π.
The given expression becomes f(y)=16sinycos3y.
Differentiating with respect to y:
f′(y)=16(cosy⋅cos3y+siny⋅3cos2y(−siny))
f′(y)=16cos2y(cos2y−3sin2y)
For maximum value, f′(y)=0:
cos2y−3sin2y=0⇒tan2y=31
Since y∈[0,2π], tany=31⇒y=6π.
Substituting y=6π in f(y):
f(6π)=16sin(6π)cos3(6π)
f(6π)=16×21×(23)3
f(6π)=8×833=33
The maximum value is 33.
Answer: 33