The brightness of some stars can fluctuate over time. Suppose that the brightness of a certain star can be modeled by the following.
\[
B(t)=8+11 \sin \left(\frac{2 \pi}{67} t\right)
\]

In this equation, $B(t)$ represents the magnitude of brighthess, and $t$ is the time (in days). Suppose we start at $t=0$ days.
During the first 67 days, when will the star's magnitude of brightness be 13 ?
Do not round any intermedlate computations, and round your answer(s) to the nearest day. (If there is more than one answer, enter additional answers with the "or" button.)
\[
1 \text { - Ddays }
\]
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Step 1 ：Given the brightness of a star can be modeled by the equation \(B(t)=8+11 \sin \left(\frac{2 \pi}{67} t\right)\), where \(B(t)\) represents the magnitude of brightness, and \(t\) is the time in days.

Step 2 ：We are asked to find when the star's magnitude of brightness will be 13 during the first 67 days, starting from \(t=0\) days.

Step 3 ：Setting \(B(t) = 13\), we get the equation \(13 = 8 + 11 \sin \left(\frac{2 \pi}{67} t\right)\).

Step 4 ：Solving for \(\sin \left(\frac{2 \pi}{67} t\right)\), we get \(\sin \left(\frac{2 \pi}{67} t\right) = \frac{5}{11}\).

Step 5 ：Using the inverse sine function, we get \(\frac{2 \pi}{67} t = \sin^{-1} \left(\frac{5}{11}\right)\).

Step 6 ：Solving for \(t\), we get \(t = \frac{67}{2 \pi} \sin^{-1} \left(\frac{5}{11}\right)\). This gives us the first solution, \(t1\).

Step 7 ：Since the sine function has a period of \(2 \pi\), there is another solution within the first 67 days. This second solution, \(t2\), is given by \(t2 = 67 - t1\).

Step 8 ：Rounding \(t1\) and \(t2\) to the nearest day, we get \(t1 = \boxed{5}\) days and \(t2 = \boxed{62}\) days.

Step 9 ：Thus, the star's magnitude of brightness will be 13 at \(t = \boxed{5}\) days or \(t = \boxed{62}\) days.