The maximum value of \(f(x) = 4\cdot \sin x\cdot \cos x\) is 2 for \(x = \frac{\pi}{4}+\pi\cdot i\), \(\forall \,i\in \mathbb{Z}\).
In this question we should use the following trigonometric identity to simplify the expression described on statement:
\(\sin 2x = 2\cdot \sin x \cdot \cos x\) (1)
Then, the expression can be rewritten as following:
\(f(x) = 2\cdot \sin 2x\) (2)
Sine is a periodic function bounded between 0 and 1. Hence, we must find the set of solutions such that sine reaches its maximum. That is to say:
\(\sin 2x = 1\)
\(2\cdot x = \sin^{-1} 1\)
\(2\cdot x = \frac{\pi}{2}+2\pi\cdot i\), \(\forall \,i\in \mathbb{Z}\)
\(x = \frac{\pi}{4}+\pi\cdot i\), \(\forall \,i\in \mathbb{Z}\)
Hence, the maximum value of \(f(x) = 4\cdot \sin x\cdot \cos x\) is 2 for \(x = \frac{\pi}{4}+\pi\cdot i\), \(\forall \,i\in \mathbb{Z}\).
We kindly invite to check this question on maxima and minima: https://brainly.com/question/12870574
