Answer
(a) $\frac{2\pi}{3}$
(b) $-\frac{\pi}{3}$
(c) $0$
(d) $\frac{2\pi}{3}$
Work Step by Step
(a) Since $cos\frac{2\pi}{3}=-\frac{1}{2}$, we have $y=arccos(-\frac{1}{2})=\frac{2\pi}{3}$
(b) Since $sin(-\frac{\pi}{3})=-\frac{\sqrt 3}{2}$, we have $y=sin^{-1}(-\frac{\sqrt 3}{2})=-\frac{\pi}{3}$
(c) Since $tan(0)=0$, we have $y=tan^{-1}(0)=0$
(d) Since $sec\frac{2\pi}{3}=-2$, we have $y=arcsec(-2)=\frac{2\pi}{3}$
