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**duaneg37** My position function is $\displaystyle x(t)=\frac{5}{12}sin(5t-.927)$

To find the time I set $\displaystyle \frac{1}{2}=sin(5t-.927) $ I know the sine of $\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{5\pi}{6}$ will give $\displaystyle \frac{1}{2}$, so I get $\displaystyle t=.29 $and $\displaystyle t=.709$. Do I use both of these times to get my answer? I found the period to be $\displaystyle T=\frac{2\pi}{5}$. I said $\displaystyle t=.29+\frac{2n\pi}{5} $ s where $\displaystyle n=0,1,2,3,...$ and $\displaystyle t=.709+\frac{2n\pi}{5}$ s where $\displaystyle n=0,1,2,3,...$ as my answer. Am I doing this right?