Let's solve the given trigonometric equation:
[tex]\[
\cos(2x) + \frac{1}{2} = 0
\][/tex]
First, isolate the cosine term:
[tex]\[
\cos(2x) = -\frac{1}{2}
\][/tex]
We need to find the values of [tex]\(2x\)[/tex] where the cosine function is [tex]\(-\frac{1}{2}\)[/tex].
In the interval [tex]\([0, 2\pi]\)[/tex], the cosine function equals [tex]\(-\frac{1}{2}\)[/tex] at:
[tex]\[
2x = \frac{2\pi}{3}, \frac{4\pi}{3}
\][/tex]
Now, solve for [tex]\(x\)[/tex] by dividing both sides by 2:
[tex]\[
x = \frac{\frac{2\pi}{3}}{2} = \frac{\pi}{3}
\][/tex]
[tex]\[
x = \frac{\frac{4\pi}{3}}{2} = \frac{2\pi}{3}
\][/tex]
Since the cosine function is periodic with period [tex]\(2\pi\)[/tex], we must include all solutions that differ by integer multiples of the period of [tex]\(x\)[/tex], which is [tex]\(\pi\)[/tex]. Therefore, we write the general solutions:
[tex]\[
x = \frac{\pi}{3} + \pi n \quad \text{for integer } n
\][/tex]
[tex]\[
x = \frac{2\pi}{3} + \pi n \quad \text{for integer } n
\][/tex]
Thus, the complete solution set to the equation [tex]\(\cos(2x) + \frac{1}{2} = 0\)[/tex] is:
[tex]\[
x = \frac{\pi}{3} + \pi n \quad \text{and} \quad x = \frac{2\pi}{3} + \pi n \quad \text{where } n \in \mathbb{Z}
\][/tex]
