A weight is attached to an elastic spring that is suspended from a ceiling. If the weight is pulled 1 inch below its rest position and released its displacement in inches after t seconds is given by x(t)=2cos(5ðt+ð/3). Find the first two times for which the displacement s 1.5 inches.

what's the problem? You have the formula:

x(t) = 2cos(5pi t + pi/3)
just solve

2cos(5pi t + pi/3) = 3/2
cos(5pi t + pi/3) = 3/4

cos .722 = .75

so,
5pi t + pi/3 = .722 or 2pi-.722=5.560
t = (.722-pi/3)/5pi = -.021 or 6.262
t = (5.560-pi/3)/5pi = .287

see the graph at

http://www.wolframalpha.com/input/?i=solve+2cos%285pi+t+%2B+pi%2F3%29+%3D+3%2F2

You need to find the next occurrence after 0.287 by noting that the period is 0.4

To find the first two times for which the displacement is 1.5 inches, we need to set the displacement function x(t) equal to 1.5 and solve for t.

The displacement function is given by x(t) = 2cos(5πt + π/3).

Setting x(t) = 1.5, we have:

1.5 = 2cos(5πt + π/3)

To solve for t, we need to isolate the cosine term:

cos(5πt + π/3) = 1.5/2
cos(5πt + π/3) = 0.75

Now, we need to find the values of t for which the cosine function equals 0.75. We can use the inverse cosine function (also known as the arccosine) to find these values.

arccos(0.75) = 5πt + π/3

To solve for t, we subtract π/3 from both sides of the equation:

arccos(0.75) - π/3 = 5πt

Next, we divide both sides by 5π:

(t) = [arccos(0.75) - π/3] / (5π)

This gives us the value of t for which the displacement is 1.5 inches. Now we can use a calculator to find the numerical value:

t ≈ [arccos(0.75) - π/3] / (5π)

To find the second time, we need to add the period of the displacement function to the first time:

t_2 = t + T

To find the period, we can use the formula T = (2π) / ω, where ω is the angular frequency.

In this case, we have ω = 5π, so the period is T = (2π) / (5π) = 2/5.

Substituting the values, we have:

t_2 = t + T
t_2 ≈ [arccos(0.75) - π/3] / (5π) + 2/5

These two expressions give us the first two times for which the displacement is 1.5 inches.