# Solve Trigonometric Problems

Remember that we’re typically looking for positive angles between 0 and \(2\pi \) so we’ll use the positive angle.An easy way to remember how to the positive angle here is to rotate one full revolution from the positive \(x\)-axis ( subtracting) \(\frac\).

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 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 \(t = 2\pi - \frac = \frac\).

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