The following equation involves a trigonometric equation in quadratic form. Solve the equation on the interval [tex]$[0, 2\pi)$[/tex].

[tex]\[ 4 \cos^2 x - 3 = 0 \][/tex]

What are the solutions in the interval [tex]$[0, 2\pi)$[/tex]? Select the correct choice below and, if necessary, fill in the answer box.

A. [tex]$x = \square$[/tex]
(Type your answer in radians. Use integers or fractions for any numbers in the expression. Type an exact answer.)

B. There is no solution.



Answer :

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]

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