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so as you can see on the title, it says finding the General Formula. So First lets take a look on the question:

Solve the equation. Give the general formula for all the solutions.

$$cos(2\Theta) = \frac{\sqrt{2}}{2}$$

Ok so you may have noticed, first off I have asked a similar question here. But No! It is not a similar question, notice in the question it says Give the general formula for all the solutions! Well of course, I don't read instructions properly (Very Bad Habit) so I complete it like the question I asked before. And I get:

$$\left\{\frac{\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{15\pi}{8} \right\}$$

.. Well all that was useless because supposedly finding the general formula is different.. The Answer Key says this is the answer: \begin{align*} \Theta & = \frac{\pi}{8} + \pi n\\ \Theta & = \frac{7\pi}{8} + \pi n \end{align*}

Well im totally confused on how to get the above answer. Help would be appreciated! Thank you. (Btw this is a Test-Review)

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Let $x = 2\theta$. Then the general solution of: $$ \cos x = \frac{\sqrt 2}{2} $$ is: $$ x = \frac{\pi}{4} + 2\pi n \quad\text{ or }\quad \frac{7\pi}{4} + 2\pi n \quad\text{ where } n \in \mathbb Z $$ Converting back from $x$ to $\theta$, we get: $$ 2\theta = \frac{\pi}{4} + 2\pi n \quad\text{ or }\quad \frac{7\pi}{4} + 2\pi n \quad\text{ where } n \in \mathbb Z $$ and so we can divide everything by two to get: $$ \theta = \frac{\pi}{8} + \pi n \quad\text{ or }\quad \frac{7\pi}{8} + \pi n \quad\text{ where } n \in \mathbb Z $$

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  • $\begingroup$ Ok i understood how to get pie/4 on the first step.. But where did you get 7pi/4 ?? $\endgroup$ – amanuel2 Apr 30 '16 at 20:57
  • $\begingroup$ @AmanuelBogale Remember that cosine is positive in the first and fourth quadrants. The fourth quadrant angle with reference angle $\pi/4$ is $2\pi - \pi/4 = 7\pi/4$. Alternatively, $\cos(-x) = \cos x$, so $\cos(-\frac{\pi}{4}) = \cos(\frac{\pi}{4})$. Note that $7\pi/4$ is coterminal with $-\pi/4$. $\endgroup$ – N. F. Taussig Apr 30 '16 at 21:03
  • $\begingroup$ Wow @N.F.Taussig ! I didnt know it applied to that! Ok thanks a bunch! $\endgroup$ – amanuel2 Apr 30 '16 at 21:21
  • $\begingroup$ @N.F.Taussig i just noticed after doing it multiple times.... isnt the final anwser for cos,sin,csc,sec suppose to be $+2\pi n$ ?? And tan,cot become $+ \pi n$ ? $\endgroup$ – amanuel2 Apr 30 '16 at 22:16
  • $\begingroup$ @AmanuelBogale What you are saying is true. However, our angle was initially $2\theta$. That is why Adriano let $x = 2\theta$. Notice that when he solved for $x$, his answers were $$x = \frac{\pi}{4} + \color{blue}{2}\pi n, n \in \mathbb{Z}$$ or $$x = \frac{7\pi}{4} + \color{blue}{2}\pi n, n \in \mathbb{Z}$$ When he replaced $x$ by $2\theta$, he had to divide each term of the right hand side of each equation by $2$, which is why he ended up with $+ \pi n, n \in \mathbb{Z}$ rather than $+ 2\pi n, n \in \mathbb{Z}$. $\endgroup$ – N. F. Taussig Apr 30 '16 at 22:21

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