How To Solve A Trigonometry Problem

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 : \(\frac,\;\frac,\; - \frac,\; - \frac\).

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We are looking for values of \(x\) so divide everything by 5 to get. Okay, now that we’ve gotten all possible solutions it’s time to find the solutions on the given interval. First, we need to do some rearranging and simplification.

\[\beginx & = \frac \frac,\quad n = 0, \pm 1, \pm 2, \ldots \\ x & = \frac \frac,\quad n = 0, \pm 1, \pm 2, \ldots \end\] Notice that we also divided the \(2\pi n\)by 5 as well! \[\begin\sin (2x) & = - \cos (2x)\\ \frac & = - 1\\ \tan \left( \right) & = - 1\end\] So, solving \(\sin (2x) = - \cos (2x)\) is the same as solving \(\tan (2x) = - 1\).

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.

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.

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.

\[\begin2\cos \left( t \right) & = \sqrt 3 \ \cos \left( t \right) & = \frac\end\] We are looking for all the values of \(t\) for which cosine will have the value of \(\frac\).

So, let’s take a look at the following unit circle.

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