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I am trying to determine the exact solutions for the angle theta, where $0 \le \theta < 2\pi$

The equation I have been given is $$10\sec\theta+2=-18$$

I am having trouble with this question as I am getting a negative which you cant do the inverse of $\cos$ of.

Any help with this would be really appreciated! Thank you very much!

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    $\begingroup$ We have $\sec\theta=-2$, so $\cos\theta=-\frac{1}{2}$. Who has cosine equal to $-\frac{1}{2}$? There are two answers between $0$ and $2\pi$. $\endgroup$ – André Nicolas Jul 22 '15 at 22:16
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    $\begingroup$ You can do inverse cosine of a negative value. It means the angle will be obtuse (i.e. Between pi/2 and pi). $\endgroup$ – John Molokach Jul 22 '15 at 22:16
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    $\begingroup$ A note on how you typed this: "where 0 is greater than or equal to theta which is greater than or equal to 2 pi" would be interpreted as $0\geq \theta \geq 2\pi$, and in particular that $0\geq 2\pi$ which is incorrect. You likely meant to say that $0$ is less than or equal to... Also, as a friendly reminder, by visiting this page, you can view a primer on how to type mathematics on this site effectively making your questions easier to read and understand. $\endgroup$ – JMoravitz Jul 22 '15 at 22:21
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Subtracting $2$ from both sides and dividing by $10$, as well as representing $\sec \theta = \frac{1}{\cos \theta}$ yields $$\frac{1}{\cos \theta} = -2 \iff \cos \theta = -\frac{1}{2}$$

This has one trivial solution $\arccos \left(-\frac{1}{2}\right) = \arccos \left(\frac{1}{2}\right) = \frac{2\pi}{3}$, can you find the other solution using a "CAST" diagram or any other method you know of?

Hint: $\cos \theta$ is negative in the second and third quadrants.

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