# How to solve $\cos(5\alpha + \pi/2) = \cos(2\alpha + \pi/8)$ for $a$?

I missed the lecture. I don't want you to solve my homework, I just want to learn how to solve equations like this one. Since I have no idea, I'll post the task I got for homework, rather than obfuscating it beyond recognition. Please give general directions on how to solve equations like this one.

$$\cos(5\alpha + \pi/2) = \cos(2\alpha + \pi/8)$$

• If you look at either the graph of cosine or its definition using the unit circle, then you can "see/conclude" that $\cos x=\cos y$, if and only if either $x=y+n\cdot2\pi$ for some integer $n$, or $x=-y+n\cdot2\pi$ for some integer $n$. Your equation is of the form "cosine of something = cosine of something else", so this can be used. You should end up with two families of solutions. It may or may not be possible to easily combine the two families, I haven't checked. Lycka till! – Jyrki Lahtonen Nov 25 '14 at 21:57

The rule for solving such a cosine equation is based on the following identity: $cosA=cosB$ implies $A=B + 2k\pi$ or $A=-B + 2k\pi$ So $5\alpha + \pi/2=2\alpha + \pi/8$ or $5\alpha + \pi/2=-(2\alpha + \pi/8)$ with $2k\pi$ Now these are equations you should be able to solve, otherwise I WILL be doing your homework :)
There are two steps to solve that question. First of all, think that if a=b, then $cos (a) = cos(b)$ So the first thing that you can do is simply to equate the two terms. The second thing you want to do is to remember that there are an infinite number of solutions for any given cosine or sine; if you rotate the radius of the unit circle all the way around, you will have a different value for the angle, but the same value for the sine and cosine. Another angle that will share sine and/or cosine depends on the starting quadrant.