What is the Coordinate of Point B in a Sinusoidal Wave?

[jpond89]

Homework Statement

Consider the sinusoidal wave in the figure, with the wave function below.

At a certain instant, let point A be at the origin and point B be the first point along the x-axis where the wave is 60.0° out of phase with point A. What is the coordinate of point B?

y = (15.0 cm) cos(0.157x - 50.3t)

and I am not allowed to post the image but here's some helpful information:
its a cosine graph, and \lambda = 40 cm, T = 0.125 s, f = 8.00 hz, k = 0.157 rads/cm

Homework Equations

Some equations that I have are:
k=2\pi/\lambda
v=\lambda*f
y=Asin(kx-\omegat)

The Attempt at a Solution

What I have tried so far is:
y=15*sin((.157)x-(50.3)t + \pi/3)
The example says " The vertical position of an element of the medium at t = 0 and x = 0 is also 15.0 cm "
The only problem is that I am really stuck and I cannot seem what to set the 60.0\circ equal to. Help would be much appreciated! Thank you!

 

[cepheid]

Everything in the argument of the cosine is the phase of the wave. So the phase is a function of both x and t. However, we're looking at this wave at a particular "instant", which means that t is constant. So, the only thing that would cause a phase change is x. So how far from the origin would the x coordinate of point B have to be in order for 0.157x - 50.3t to differ from 0.157*(0) - 50.3t by 60 degrees, given that t is constant?

 

[jpond89]

Would x = 6.37 * \pi because that would make it (\pi/3 -50.3t)

 

[cepheid]

I think you might be off by a factor of 3. Check your work again.