In a calculus class we are often more interested in only the solutions to a trig equation that fall in a certain interval.
Likewise, the angle in the fourth quadrant will \(\frac\) below the \(x\)-axis.
So, we could use \( - \frac\) or \(2\pi - \frac = \frac\).
We first need to get the trig function on one side by itself.
To do this all we need to do is divide both sides by 2.
Therefore, since there isn’t anything in this problem (contrast this with the next problem) to tell us which is the correct solution we will need to list ALL possible solutions. Recall from the previous section and you’ll see there that we used \[\frac 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \] to represent all the possible angles that can end at the same location on the unit circle, angles that end at \(\frac\).
Remember that all this says is that we start at \(\frac\) then rotate around in the counter-clockwise direction (\(n\) is positive) or clockwise direction (\(n\) is negative) for \(n\) complete rotations.The same thing can be done for the second solution.So, all together the complete solution to this problem is \[\begin\frac 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \ \frac 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \end\] As a final thought, notice that we can get \( - \frac\) by using \(n = - 1\) in the second solution.So, the solutions are : \(\frac,\;\frac,\; - \frac,\; - \frac\).This problem is very similar to the other problems in this section with a very important difference.Now we come to the very important difference between this problem and the previous problems in this section.The solution is NOT \[\beginx & = \frac 2\pi n,\quad n = 0, \pm 1, \pm 2, \ldots \ x & = \frac 2\pi n,\quad n = 0, \pm 1, \pm 2, \ldots \end\] This is not the set of solutions because we are NOT looking for values of \(x\) for which \(\sin \left( x \right) = - \frac\), but instead we are looking for values of \(x\) for which \(\sin \left( \right) = - \frac\).One way to remember how to get the positive form of the second angle is to think of making one full revolution from the positive \(x\)-axis ( subtracting) \(\frac\). As the discussion about finding the second angle has shown there are many ways to write any given angle on the unit circle.Sometimes it will be \( - \frac\) that we want for the solution and sometimes we will want both (or neither) of the listed angles.The angle in the first quadrant makes an angle of \(\frac\) with the positive \(x\)-axis, then so must the angle in the fourth quadrant. We could use \( - \frac\), but again, it’s more common to use positive angles.To get a positive angle all we need to do is use the fact that the angle is \(\frac\) with the positive \(x\)-axis (as noted above) and a positive angle will be \(t = 2\pi - \frac = \frac\).