Let's solve the equation \(4 \cos^2 x - 3 = 0\) within the interval \([0, 2\pi)\).
1. Isolate \(\cos^2 x\):
[tex]\[
4 \cos^2 x - 3 = 0
\][/tex]
Add \(3\) to both sides:
[tex]\[
4 \cos^2 x = 3
\][/tex]
Divide both sides by \(4\):
[tex]\[
\cos^2 x = \frac{3}{4}
\][/tex]
2. Take the square root of both sides:
[tex]\[
\cos x = \pm \sqrt{\frac{3}{4}}
\][/tex]
Since \(\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\), we have:
[tex]\[
\cos x = \pm \frac{\sqrt{3}}{2}
\][/tex]
3. Determine the values of \(x\) within \([0, 2\pi)\):
- For \(\cos x = \frac{\sqrt{3}}{2}\):
- \(x = \frac{\pi}{6}\)
- This cosine value appears again in the fourth quadrant at \(x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}\)
- For \(\cos x = -\frac{\sqrt{3}}{2}\):
- This cosine value appears in the second quadrant at \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\)
- It also appears in the third quadrant at \(x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\)
Therefore, the solutions to the equation \(4 \cos^2 x - 3 = 0\) in the interval \([0, 2\pi)\) are:
[tex]\[
x = \left\{\frac{\pi}{6}, \frac{11\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}\right\}
\][/tex]
So the correct choice is:
A. [tex]\( x = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6} \)[/tex]
