Find the roots of $f(x) = 2\cos(x)$

Question

From a book I know that the roots of $\cos(x)$ are: $$\left[\ldots,-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\ldots\right]$$

What are the roots of $2\cos(x)$?

I have the answer here: $$\left[\ldots,-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\ldots\right]$$

I just don't see the connection or how this was arrived at.

Why are those the roots of $2\cos(x)$?

Answer 1

If $x$ is a real number such that $\cos(x) = 0$, what is the value of $2\cos(x)$?

On the other hand, if $x$ is a real number such that $2\cos(x) = 0$, what is the value of $\cos(x)$?

Now can you see why the roots of $\cos(x)$ and the roots of $2\cos(x)$ are the same?

Answer 2

Hint:

If $2\cdot \cos(x)=0$, then either:

• $2=0$
• $\cos(x)=0$