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What is the phase constant δ in x = Acos(ωt + δ) if the position of the oscillating particle at time t = 0 is the following?

In other words -
What is the phase constant Delta in x = A*cos(omega*t + Delta) if the position of the oscillating particle at time t = 0 is the following?

(a) x0 = 0
(Enter your answers from smallest to largest.)

(b) x0 = −A

(c) x0 = A

(d) x0 = A/2

x0 = x at t = 0
Edit
Added Tue, 09 Jun '15
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Because time equals 0 in all cases, the system we are solving becomes

x(t) = Acos(ωt + δ)
x(0) = x0 = A*cos(δ)

x0 = A*cos(δ)

From here we just need a basic understanding of trig.

(a) x0 = 0
What two values of δ does A*cos(δ) = 0?

A*cos(δ) = 0
>> cos(δ) = 0
>> δ = arccos(0)
δ = arccos(0) = Pi/2 and 3Pi/2.

In other words,
cos(δ) = 0 at Pi/2 and 3Pi/2. Thus A*cos(δ) will = 0 at Pi/2 and 3Pi/2.

x0 = 0 at
δ = Pi/2 radians = 1.57 radians
and
δ = 3Pi/2 radians = 4.71 radians

(b) x0 = −A
What value of δ does A*cos(δ) = -A?

A*cos(δ) = -A
>> cos(δ) = -1
>> δ = arccos(-1)
δ = arccos(-1) = Pi

In other words,
cos(δ) = -1 at Pi. Thus A*cos(δ) will = -A at Pi.

x0 = -A at
δ = Pi radians = 3.14 radians

(c) x0 = A
What value of δ does A*cos(δ) = A?

A*cos(δ) = A
>> cos(δ) = 1
>> δ = arccos(1)
δ = arccos(1) = 0.

x0 = A at
δ = 0.

(d) x0 = A/2
What value of δ does A*cos(δ) = A/2?

A*cos(δ) = A/2
>> cos(δ) = 1/2
>> δ = arccos(1/2)
δ = arccos(1/2) = Pi / 3

x0 = A/2 at
δ = Pi/3 radians = 1.046 radians
Edit
Added Tue, 09 Jun '15
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