A spring is oscillating vertically according to the equation y = 3cos(2t-pi), where y is the displacement in feet and t is time in seconds. At what time will thespring be 2 ft. high? Round your answer to the nearest tenth.

Answer: 25.7 seconds
Explanation
Equation
[tex]y=3\cos(2t-\pi)[/tex]where y is the displacement in feet and t is time in seconds.
We are asked to find t when y = 2ft, meaning that we have to solve the equation for t:
0. Replacing the information given by setting the equation to 2:
[tex]3\cos(2t-\pi)=2[/tex]2. Dividing both sides of the equation against 3 and simplifying:
[tex]\frac{3\cos(2t-\pi)}{3}=\frac{2}{3}[/tex][tex]\cos(2t-\pi)=\frac{2}{3}[/tex]2. Applying secant (cos⁻¹) to both sides:
[tex]\cos^{-1}(\cos(2t-\pi)=\cos^{-1}(\frac{2}{3})[/tex][tex](2t-\pi)=\cos^{-1}(\frac{2}{3})[/tex]3. Adding π to both sides of the equation:
[tex]2t-\pi+\pi=\cos^{-1}(\frac{2}{3})+\pi[/tex][tex]2t=\cos^{-1}(\frac{2}{3})+\pi[/tex]4. Dividing both sides of the equation against two:
[tex]\frac{2t}{2}=\frac{\cos^{-1}(\frac{2}{3})+\pi}{2}[/tex][tex]t=\frac{\cos^{-1}(\frac{2}{3})+\pi}{2}[/tex]5. Simplifying we get:
[tex]t\approx\frac{51.33}{2}=25.67[/tex]