In my high school math class I'm currently working on solving trigonometric equations.
Solve for all possible values of x:
$5 = 30\cos(2x + 1) - 5 $
My teacher (on some examples posted online) achieved this result.
$x = \frac{\arctan(\frac{1}{3}) \pm 180 + 1}{2}$
My question is: Why is arctan being used? Is there a relationship between arctan and cos that I'm missing? Why isn't arccos being used instead?
Also the plus-minus 180. I think it has something to do with the range of the cos function but I'm not exactly sure either.
Sorry for any formatting issues, first time poster.
We have: $$5=30\cos(2x+1)-5$$ $$\to \frac 13 =\cos(2x+1)$$ $$\to 2x+1 =\arccos(\frac 13)$$ $$\to x=\frac12(\arccos(\frac13)-1)$$
Now, notice that: $$\cos(x)=\cos(x+360)=\cos(x-360)=\cos(x+720)=...$$
Because $\cos(x)$ is a periodic function with period $360^0$
Because of this, $$\arccos(\frac 13)=I\pm360k$$
Also, because $\cos(x)=\cos(-x)$, we have $-\arccos(\frac 13)$ fittingas well.
For any integer $k$, where $I\approx70.527$ is the initial value of $\arccos(\frac13)$
That's why the $\pm 180$ bit came in to the solution, even if it's slightly off.