To get the first angle here let’s recall the definition of tangent in terms of sine and cosine.

\[\tan \left( {\frac{t}{4}} \right) = \frac{{\sin \left( {{\textstyle{t \over 4}}} \right)}}{{\cos \left( {{\textstyle{t \over 4}}} \right)}} = - \frac{1}{{\sqrt 3 }}\]

Now, because of the section we’re in, we know that the angle must be one of the “standard” angles and from a quick look at a unit circle (shown below) we know that for \(\frac{\pi }{6}\) we will have,

\[\frac{\sin \left( \tfrac{\pi }{6} \right)}{\cos \left( \tfrac{\pi }{6} \right)}=\frac{{}^{1}/{}_{2}}{{}^{\sqrt{3}}/{}_{2}}=\frac{1}{\sqrt{3}}\]

So, if we had a positive value on the tangent we’d have the first angle. We do have a negative value however, but this work will allow us to get the two angles we’re after. Because the value is negative this simply means that the sine and cosine must have the same values that they have for \(\frac{\pi }{6}\) except that one must be positive and the other must be negative. This means that the angles that we’re after must be in the second and fourth quadrants. Here is a unit circle for this situation.

By basic symmetry we can see that the terminal line for the angle in the second quadrant must form an angle of \(\frac{\pi }{6}\) with the negative \(x\)-axis and the terminal line in the fourth quadrant must form an angle of \(\frac{\pi }{6}\) with the positive \(x\)-axis as shown above. The angle in the second quadrant will then be : \(\pi - \frac{\pi }{6} = \frac{{5\pi }}{6}\) while the angle in the fourth quadrant will be \(2\pi - \frac{\pi }{6} = \frac{{11\pi }}{6}\).

Note that you don’t really need a positive angle for the second one. If you wanted to you could just have easily used \( - \frac{\pi }{6}\) for the second angle. There is nothing wrong with this and you’ll get the same solutions in the end. The reason we chose to go with the positive angle is simply to avoid inadvertently losing the minus sign on the second solution at some point in the future. That kind of mistake is easy to make on occasion and by using positive angles here we will not need to worry about making it.

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,

\[\frac{t}{4} = \frac{{5\pi }}{6} + 2\pi n \hspace{0.5in} {\rm{OR }} \hspace{0.5in} \frac{t}{4} = \frac{{11\pi }}{6} + 2\pi n \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]

Finally, to get all the solutions to the equation all we need to do is multiply both sides by 4.

\[t = \frac{{10\pi }}{3} + 8\pi n \hspace{0.5in} {\rm{OR }} \hspace{0.5in} t = \frac{{22\pi }}{3} + 8\pi n \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \] Show Step 4

Note that because at least some of the solutions will have a denominator of 3 it will probably be convenient to also have the interval written in terms of fractions with denominators of 3. Doing this will make it much easier to identify solutions that fall inside the interval so,

\[\left[ {0,4\pi } \right] = \left[ {0,\frac{{12\pi }}{3}} \right]\]

With the interval written in this form, if our potential solutions have a denominator of 3, all we need to do is compare numerators. As long as the numerators are positive and less than \(12\pi \) we’ll know that the solution is in the interval.

Also, in order to quickly determine the solution for particular values of \(n\) it will be much easier to have both fractions in the solutions have denominators of 3. So the solutions, written in this form, are.

\[t = \frac{{10\pi }}{3} + \frac{{24\pi n}}{3} \hspace{0.5in} {\rm{OR }} \hspace{0.5in} t = \frac{{22\pi }}{3} + \frac{{24\pi n}}{3} \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]

Now let’s find all the solutions. First notice that, in this case, if we plug in negative values of \(n\) we will get negative solutions and these will not be in the interval and so there is no reason to even try these. Next, notice that for any positive \(n\) we will be adding \(\frac{{24\pi }}{3}\) onto a positive quantity and so are guaranteed to be greater than \(\frac{{12\pi }}{3}\) and so will out of the given interval. This leaves \(n = 0\) and for this one we can notice that the only solution that will fall in the given interval is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{10\pi }}{3}}}\]

Before leaving this problem let’s note that on occasion we will only get a single solution out of all the possible solutions that will fall in the given interval. So, don’t get excited about it if this should happen.