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.
$$\cos(5\alpha + \pi/2) = \cos(2\alpha + \pi/8)$$
 A: 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 :)
A: 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.
A: Use Cos(C) - Cos(D) formula.
U can also solve by graphical method. You know the graph of Cosx then apply transformation to draw LHS and RHS intersection point will be the solution
