# 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?

• This is phrased a bit too much like a homework problem for comfort. Could you tell us at what points you ran into difficulty with it? – Michael Hardy Oct 28 '14 at 17:31
• Question edited. – Kiara Oct 29 '14 at 4:21
• Wow! This is a perfect example of how people should edit their questions in response to requests for more detail. I, for one, have voted to reopen. :) (It requires 5 votes, I believe, to reopen.) – apnorton Oct 29 '14 at 4:22
• @anorton thank you. I was busy trying to solve it by myself so couldn't write so much. At the end I was unsuccessful and had to edit it! :( – Kiara Oct 29 '14 at 4:23

\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}

• Your answers are right! But I don't understand few things here. How does $360^0k$ come here? – Kiara Oct 29 '14 at 4:25
• Consider that trig functions would be periodic about that many degrees. For example, if the initial right hand side had a value of 1, would you just look at 0 degrees or would you consider -720, -360, 360, 720, and others that may exist? – JB King Oct 29 '14 at 5:20
• Exactly like JB King said. If $cos(x) = a$, then $cos(x + 360^\circ)=a$. – Pradu Oct 30 '14 at 21:18

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.