Because we’re dealing with sine in this problem and we know that the \(y\)-axis represents sine on a unit circle we’re looking for angles that will have a \(y\) coordinate of \({\textstyle{1 \over 2}}\). This means we’ll have an angle in the first quadrant and an angle in the second quadrant (that we can use the angle in the first quadrant to find). Here is a unit circle for this situation.

Clearly the angle in the first quadrant is \(\frac{\pi }{6}\) and by some basic symmetry we can see that the terminal line for the second angle must form an angle of \(\frac{\pi }{6}\) with the negative \(x\)-axis as shown above and so it will be : \(\pi - \frac{\pi }{6} = \frac{{5\pi }}{6}\).

Show Step 3

From the discussion in the notes for this section we know that once we have these two angles we can get all possible angles by simply adding “\( + \,2\pi n\) for \(n = 0, \pm 1, \pm 2, \ldots \)” onto each of these.

This then means that we must have,

\[3t = \frac{\pi }{6} + 2\pi n \hspace{0.5in}{\rm{OR }}\hspace{0.5in}3t = \frac{{5\pi }}{6} + 2\pi n \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]

Finally, to get all the solutions to the equation all we need to do is divide both sides by 3.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{t = \frac{\pi }{{18}} + \frac{{2\pi n}}{3} \hspace{0.5in} {\rm{OR }} \hspace{0.5in} t = \frac{{5\pi }}{{18}} + \frac{{2\pi n}}{3} \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots }}\]