How to find all solutions to $\cos(x) = a$?

Question

I have the following question at hand:

Find all extrema of the function $f(x) = x - 2\sin(x+ \frac{\pi}{4})$. That amounts to solving $\cos(x + \frac{\pi}{4}) = \frac{1}{2}$. But simply using $\arccos(0.5) - \frac{\pi}{4}$ yields only one solution; I need all of them. How can I find them? Thanks in advance.

Answer 1

Alright I found a solution: Because the first two solutions are $\pm arccos(0.5) - \frac{\pi}{4}$, the period $2\pi$ dictates all the other ones:

$\frac{\pi}{12} + 2k\pi$ and $-\frac{7\pi}{12} + 2k\pi$.

Answer 2

Both the sine and cosine functions are periodic, with period $2\pi.$ This is usually seen in a first course in precalculus or trigonometry.

Take the unit circle (the circle of radius $1$ centered at the origin). A point on the circumference of the circle has coordinates $(\cos \theta, \sin \theta)$, where $\theta$ is the angle measured from the positive $x$-axis in the counterclockwise direction. If you start at any point and do one full rotation, you will return to that point. Recall that one full rotation is $2\pi$ radians.

Now suppose we draw a vertical line that intersects the circle twice. The angle formed by the center and the two points of intersection is a right angle - $\pi$ radians. So if we need to solve $\cos x = a$, what do we do? Take two cases, and account for periodicity.

A simple example - all solutions to $\cos x = \frac{1}{2}$ are angles of the form $\frac{\pi}{3} + 2\pi k$ and $-\frac{\pi}{3} + 2\pi k$ for integers $k$. Why? The principal range of the arccosine function is the interval $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}.$ This is to ensure we actually have an inverse function.

Now you can apply this reasoning to $\cos \left( x + \frac{\pi}{4} \right) = \frac{1}{2}$ by taking two cases, and subtracting from the "representative" angles I've given above. See chapters 5-7 of OpenStax Precalculus for more.