Likewise, the angle in the fourth quadrant will \(\frac\) below the \(x\)-axis.

So, we could use \( - \frac\) or \(2\pi - \frac = \frac\).

We’ll start this problem in exactly the same way as we did in the first example. \[\begin2\sin (5x) & = - \sqrt 3 \\ \sin (5x) & = \frac\end\] We are looking for angles that will give \( - \frac\) out of the sine function. Now, there are no angles in the first quadrant for which sine has a value of \( - \frac\). Given this we now know that the angle in the third quadrant will be \(\frac\) below the \(x\)-axis or \(\pi \frac = \frac\).

However, there are two angles in the lower half of the unit circle for which sine will have a value of \( - \frac\). An easy way to remember this is to notice that we’ll rotate half a revolution from the positive \(x\)-axis to get to the negative \(x\)-axis then add on \(\frac\) to reach the angle we are looking for.

In this section we will take a look at solving trig equations.

This is something that you will be asked to do on a fairly regular basis in many classes.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.From quick inspection we can see that \(t = \frac\) is a solution.However, as we have shown on the unit circle there is another angle which will also be a solution. When we look for these angles we typically want angles that lie between 0 and \(2\pi \).Yes, the sine, on the first period, takes on the value of I've done the algebra; that is, I've done the factoring and then I've solved each of the two factor-related equations. So now I can do the trig; namely, solving those two resulting trigonometric equations, using what I've memorized about the cosine wave.From the first equation, I get: However (and this is important! Due to the nature of the mathematics on this site it is best views in landscape mode.If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.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.

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