Since d represents the ball's distance from the rest position, the position of 2 inches above the rest position is described by d=2.
To find the values of t for which the ball is located at d=2, replace d=2 and solve for t:
[tex]\begin{gathered} d=-4\cos (\frac{\pi}{3}t) \\ \Rightarrow2=-4\cos (\frac{\pi}{3}t) \\ \Rightarrow\frac{2}{-4}=\frac{-4\cos (\frac{\pi}{3}t)}{-4} \\ \Rightarrow-\frac{1}{2}=\cos (\frac{\pi}{3}t) \\ \Rightarrow\cos (\frac{\pi}{3}t)=-\frac{1}{2} \end{gathered}[/tex]
Take the inverse cosine of both members of the equation:
[tex]\begin{gathered} \Rightarrow\cos ^{-1}(\cos (\frac{\pi}{3}t))=\cos ^{-1}(-\frac{1}{2}) \\ \Rightarrow\frac{\pi}{3}t=\pm\frac{2}{3}\pi+2\pi k \end{gathered}[/tex]
Where k is any integer number.
Multiply both sides by 3 and divide both sides by π:
[tex]\begin{gathered} \frac{3}{\pi}\times\frac{\pi}{3}t=\frac{3}{\pi}\times(\pm\frac{2}{3}\pi+2\pi k) \\ \Rightarrow t=\pm2+6k \end{gathered}[/tex]
Therefore, all the values of t for which the ball is 2 inches above its rest position are described by the expression:
[tex]\begin{gathered} t=6k\pm2 \\ k\in\Z \end{gathered}[/tex]
