An oscillator has an amplitude of 3.2. At this instant the displacement of the oscillator is 1.4. What are the two possible phases of the oscillator at this instant?

so i used the equation x=A cos (omega(t) + phase zero)

which looks like
1.4=3.2 cos phase
.4375=cos phase
so than i take
cos^-1 of .4375 to get the phase..is this right?

That looks good.

Remmeber there will be two phases that have a positive cosine value.

how do i figure that out?

arccos(0.4375)=+-1.118 radians.

cos(-1.118 radians) = cos( (2*pi) - 1.118 )
=cos(5.1652 radians)

either of these will give a cosine value of +0.4375

Yes, your thinking is correct. To find the phase of the oscillator at the given instant, you can use the equation x = A cos(ωt + φ), where x is the displacement of the oscillator, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase.

In this case, you have the displacement x = 1.4 and the amplitude A = 3.2. So, the equation becomes 1.4 = 3.2 cos(ωt + φ).

To solve for the phase φ, you can rearrange the equation like this: cos(ωt + φ) = 1.4 / 3.2.

Now, you can take the inverse cosine (cos⁻¹) of both sides of the equation to isolate the phase φ. So, cos⁻¹(cos(ωt + φ)) = cos⁻¹(1.4 / 3.2).

The inverse cosine function undoes the cosine function, giving you the angle whose cosine is the given value. Therefore, cos⁻¹(1.4 / 3.2) will give you the phase φ.

By solving this equation, you will obtain the value of the phase at the given instant.