Answer
$a.\displaystyle \quad \frac{\pi}{3}$
$b.\displaystyle \quad \frac{3\pi}{4}$
$c.\displaystyle \quad \frac{\pi}{6}$
Work Step by Step
$y=\cos^{-1}x$ is the number in $[0, \pi]$ for which $\cos y=x.$
Use reference angles in the 1st quadrant,
keeping in mind that $\cos(\pi-t)=-\cos t$
$\left\{\begin{array}{llll}
\cos\frac{\pi}{3}=\frac{1}{2} & \Rightarrow & & \Rightarrow\cos^{-1}(\frac{1}{2})=\dfrac{\pi}{3}\\
& & & \\
\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}} & \Rightarrow & \cos(-\frac{\pi}{4})=-\frac{1}{\sqrt{2}} & \Rightarrow\cos^{-1}(-\frac{1}{\sqrt{2}})=\pi-\frac{\pi}{4}=\dfrac{3\pi}{4}\\
& & & \\
\cos\frac{\pi}{6}=\frac{ \sqrt{3}}{2} & \Rightarrow & & \Rightarrow\cos^{-1}(\frac{\sqrt{3}}{2})=\dfrac{\pi}{6}
\end{array}\right.$
