Answer
(a) See the graph.
(b) Quadrant I.
(c) $\frac{20\pi}{9}$ and $-\frac{16\pi}{9}$
Work Step by Step
(b)
$0\lt\frac{2\pi}{9}\lt\frac{\pi}{2}$, therefore, it lies in quadrant I.
(c)
$\frac{2\pi}{9}+2\pi=\frac{20\pi}{9}$
$\frac{2\pi}{9}-2\pi=-\frac{16\pi}{9}$
