Có \(F\left( x \right) = \int {f\left( x \right)dx} = \int {\cos 2xdx} = \frac{1}{2}\sin 2x + C.\)

Vì \(F\left( \pi \right) = 1\) nên \(\frac{1}{2}\sin 2\pi + C = 1 \Leftrightarrow C = 1\).

Do đó \(F\left( x \right) = \frac{1}{2}\sin 2x + 1\).

Phương trình \(F\left( x \right) = 1\)\( \Leftrightarrow \frac{1}{2}\sin 2x + 1 = 1\)\( \Leftrightarrow \sin 2x = 0\)\( \Leftrightarrow 2x = k\pi \)\( \Leftrightarrow x = k\frac{\pi }{2},k \in \mathbb{Z}\).

Vì \(x \in \left[ {0;3\pi } \right]\) nên \(x = 0;x = \frac{\pi }{2};x = \pi ;x = \frac{{3\pi }}{2};x = 2\pi ;x = \frac{{5\pi }}{2},x = 3\pi \).