To find the interval that makes the inequality [tex]\(4 \cos(2x) \geq 2\)[/tex] true on the domain [tex]\(0 \leq x < 2\pi\)[/tex], we will proceed with the following steps:
1. Simplify the Inequality:
[tex]\[4 \cos(2x) \geq 2\][/tex]
Dividing both sides by 2:
[tex]\[\cos(2x) \geq \frac{1}{2}\][/tex]
2. Determine the Values of [tex]\(2x\)[/tex] that Satisfy the Inequality:
We need the values of [tex]\(2x\)[/tex] where [tex]\(\cos(2x) \geq \frac{1}{2}\)[/tex]. The cosine function achieves [tex]\(\frac{1}{2}\)[/tex] at:
[tex]\[\cos(2x) = \frac{1}{2} \text{ at } 2x = \frac{\pi}{3} + 2k\pi \text{ or } 2x = -\frac{\pi}{3} + 2k\pi\][/tex]
for any integer [tex]\(k\)[/tex].
Next, we determine the intervals where [tex]\(\cos(2x) \geq \frac{1}{2}\)[/tex]. The general solutions for [tex]\(\cos(2x) \geq 1/2\)[/tex] are in the intervals:
[tex]\[\frac{\pi}{3} + 2k\pi \leq 2x \leq 5\pi/3 + 2k\pi\][/tex]
3. Solve for [tex]\(x\)[/tex] in the Domain [tex]\(0 \leq x < 2\pi\)[/tex]:
- For [tex]\(k=0\)[/tex]:
[tex]\[\frac{\pi}{3} \leq 2x \leq \frac{5\pi}{3}\][/tex]
Solving for [tex]\(x\)[/tex], divide by 2:
[tex]\[\frac{\pi}{6} \leq x \leq \frac{5\pi}{6}\][/tex]
- For [tex]\(k=1\)[/tex]:
Add [tex]\(2\pi\)[/tex] (which corresponds to one full rotation):
[tex]\[2\pi + \frac{\pi}{3} \leq 2x \leq 2\pi + \frac{5\pi}{3}\][/tex]
[tex]\[\frac{7\pi}{3} \leq 2x \leq \frac{11\pi}{3}\][/tex]
Solving for [tex]\(x\)[/tex], divide by 2:
[tex]\[\frac{7\pi}{6} \leq x \leq \frac{11\pi}{6}\][/tex]
4. Combine the Intervals:
Therefore, the x-intervals where [tex]\(\cos(2x) \geq 1/2\)[/tex] in [tex]\(0 \leq x < 2\pi\)[/tex] are:
[tex]\[\left[\frac{\pi}{6}, \frac{5\pi}{6}\right] \cup \left[\frac{7\pi}{6}, \frac{11\pi}{6}\right]\][/tex]
Hence, the correct interval that makes the inequality true is:
[tex]\[
\left[\frac{\pi}{6}, \frac{5\pi}{6}\right] \cup \left[\frac{7\pi}{6}, \frac{11\pi}{6}\right]
\][/tex]
