What is the phase of the simple harmonic motion at t = 10.0 s?

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MrAlt
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Homework Statement



The function x = (2.0 m) cos[(6π rad/s)t + π/2 rad] gives the simple harmonic motion of a body. Find the following values at t = 10.0 s.

I am having trouble finding the phase of the motion, The book gives an answer of 190 rad, I am not sure how they got that

Homework Equations



V(t) = -12[itex]\pi[/itex]Sin(6[itex]\pi[/itex]t+([itex]\pi[/itex]/2))
a(t) = -12[itex]\pi[/itex](6[itex]\pi[/itex])Cos(6[itex]\pi[/itex]t+([itex]\pi[/itex]/2))


The Attempt at a Solution



The displacement and acceleration are zero, but for some reason when i plug them into my calculator (it is set to radians) i keep getting answers other than zero, any suggestions?
 
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Just found out what i was missing, and feel like a complete fool, sorry to bother you all with this, Thank You