# How To Solve A Trigonometry Problem

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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.

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

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

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.

## How To Solve A Trigonometry Problem Research Papers On Grief And Loss

Sometimes it will be $$- \frac$$ that we want for the solution and sometimes we will want both (or neither) of the listed angles.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. 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.In a calculus class we are often more interested in only the solutions to a trig equation that fall in a certain interval.The first step in this kind of problem is to find all possible solutions. $\begin\frac 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \ \frac 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \end$ Now, to find the solutions in the interval all we need to do is start picking values of $$n$$, plugging them in and getting the solutions that will fall into the interval that we’ve been given. $\begin& \frac 2\pi \,\left( 0 \right)\, = \frac use $$n = - 2$$) we will once again be outside of the interval so we’ve found all the possible solutions that lie inside the interval $$[ - 2\pi ,2\pi ]$$.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 $$\sin \left( \right) = - \frac$$.Note the difference in the arguments of the sine function! This makes all the difference in the world in finding the solution! $\beginx & = \frac \frac = \frac 2\pi \end$ Okay, so we finally got past the right endpoint of our interval so we don’t need any more positive n.Therefore, the set of solutions is $\begin5x & = \frac 2\pi n,\quad n = 0, \pm 1, \pm 2, \ldots \ 5x & = \frac 2\pi n,\quad n = 0, \pm 1, \pm 2, \ldots \end$ Well, actually, that’s not quite the solution. $\beginx & = \frac \frac = \frac = \frac & \hspace & \Rightarrow \hspace\sin \left( \right) = \sin \left( \right) = - \frac\ x & = \frac \frac = \frac & \hspace & \Rightarrow \hspace\sin \left( \right) = \sin \left( \right) = - \frac\end$ We’ll leave it to you to verify our work showing they are solutions. If you didn’t divide the $$2\pi n$$ by 5 you would have missed these solutions! Now let’s take a look at the negative $$n$$ and see what we’ve got. \[\beginx & = \frac \frac = - \frac This problem is a little different from the previous ones.

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