Solve the equation cos(x) + cos(3x) = 0

Let's solve the equation cos(x) + cos(3x) = 0.

We can use the sum-to-product formula for cosines, which states that:

$$cos(a) + cos(b) = 2 \cdot cos(\frac{a+b}{2}) \cdot cos(\frac{a-b}{2})$$

In our case, a = x and b = 3x. Therefore, we have:

$$cos(x) + cos(3x) = 2 \cdot cos(\frac{x+3x}{2}) \cdot cos(\frac{x-3x}{2}) = 0$$

Simplify the arguments of the cosines:

$$2 \cdot cos(\frac{4x}{2}) \cdot cos(\frac{-2x}{2}) = 0$$

$$2 \cdot cos(2x) \cdot cos(-x) = 0$$

Since cosine is an even function, cos(-x) = cos(x). Thus, we have:

$$2 \cdot cos(2x) \cdot cos(x) = 0$$

This equation is satisfied if either cos(2x) = 0 or cos(x) = 0.

Case 1: cos(x) = 0

$$x = \frac{\pi}{2} + \pi k, \quad k \in \mathbb{Z}$$

Case 2: cos(2x) = 0

$$2x = \frac{\pi}{2} + \pi n, \quad n \in \mathbb{Z}$$

$$x = \frac{\pi}{4} + \frac{\pi}{2} n, \quad n \in \mathbb{Z}$$

So, the solutions are:

$$x = \frac{\pi}{2} + \pi k, \quad k \in \mathbb{Z}$$

and

$$x = \frac{\pi}{4} + \frac{\pi}{2} n, \quad n \in \mathbb{Z}$$

We can combine these solutions into a more compact form. Notice that when n is odd, say n = 2k+1, then

$$x = \frac{\pi}{4} + \frac{\pi}{2}(2k+1) = \frac{\pi}{4} + \pi k + \frac{\pi}{2} = \frac{3\pi}{4} + \pi k$$

And when n is even, say n = 2k, then

$$x = \frac{\pi}{4} + \frac{\pi}{2}(2k) = \frac{\pi}{4} + \pi k$$

So the general solution is:

$$x = \frac{\pi}{4} + \frac{\pi}{2} k, \quad k \in \mathbb{Z}$$

Answer: $$x = \frac{\pi}{4} + \frac{\pi}{2} k, \quad k \in \mathbb{Z}$$