Solve equation for $0^\circ < x < 360^\circ$ Solve the following equation for $0^\circ < x < 360^\circ$ 
$$\cos(2x - 15^\circ) = -0.145$$
By finding out the cos inverse, I get $81.7^\circ$. Because $-0.145$ is negative, it lies on the 2nd and 3rd quadrant. That is we can find $x$ by $(180 + \theta)$ and $(180 - \theta)$.
So this was what I've thought:
$$2x-15^\circ = 81.7^\circ$$
From here according to me, $$x=(81.7^\circ + 15^\circ)/2$$ gives $$x=48.4$$ which is not the answer in my book.
Then comes $$2x-15^\circ=(180+81.7^\circ),(180-81.7^\circ)$$
From here I find $$x=138.35^\circ, 56.7^\circ$$ these answers are correct.
Now, the first answer $48.4$ is wrong, and there is one more answer in my book, i.e, there are $4$ values in the answer, where I got $3$ answers and $1$ is wrong.
Could someone help help me to find the two other answers?
 A: \begin{align}
\cos( 2x - 15^\circ ) &= -0.145 \\
2x - 15^\circ &= \cos^{-1}(-0.145) + 360^\circ k \quad\quad \ k\in \ldots -2, -1, 0, 1, 2, \ldots  \\
x  &= \frac{1}{2} \left( 15^\circ + \cos^{-1}(-0.145) + 360^\circ k \right) \quad\quad \ k\in \ldots -2, -1, 0, 1, 2, \ldots  \\
x  &= \frac{1}{2} \left( 15^\circ \pm 98.3372791889057^\circ + 360^\circ k \right) \quad\quad \ k\in \ldots -2, -1, 0, 1, 2, \ldots  \\
x &=  56.6686395944528^\circ, 138.331360405547^\circ, 236.668639594453^\circ, 318.331360405547^\circ
\end{align}
A: For this post I assume you're using degrees and not something else like radians.
You could try using the inverse cosine function which is called $\arccos$ or $\cos^{-1}$ depending on who you ask. Below I'll be using $\arccos$ but if you normally see $\cos^{-1}$ then just know that they are the same.

Here's an example of using it:
$$\cos(x)=0.5$$
I take the function $\arccos$ on both sides
$$\arccos(\cos(x))=\arccos(0.5)$$
You can calculate $\arccos(x)$ on your calculator, make sure it's set on degrees mode.
$$x=60^{\circ}$$

This works because $\cos(x)$ and $\arccos(x)$ cancel each other just like $\sqrt{x}$ and $x^2$ cancels each other.
After you've used the function you have a much simpler equation which you probably can solve yourself.
