Question

solve the equation on the interval ( 0 leq \theta<2 pi ).
( 4 cos ^{2} \theta=3 )
what are the solutions in the interval ( 0 leq \theta<2 pi ) ? select the correct choice and fill in any answer boxes in your choice below.
a. the solution set is {}
(simplify your answer. type an exact answer, using ( pi ) as needed. type your answer in radians. use integers or fractions for any numbers in the expression as needed.)
b. there is no solution.

Explanation:

Step1: Solve for \(\cos\theta\)

Given \(4\cos^{2}\theta = 3\), divide both sides by \(4\):
\(\cos^{2}\theta=\frac{3}{4}\).
Take the square - root of both sides: \(\cos\theta=\pm\frac{\sqrt{3}}{2}\).

Step2: Find \(\theta\) values

When \(\cos\theta=\frac{\sqrt{3}}{2}\), \(\theta = \frac{\pi}{6}+2k\pi\) or \(\theta=\frac{11\pi}{6}+2k\pi\) (\(k\in\mathbb{Z}\)).
In the interval \(0\leq\theta < 2\pi\), when \(k = 0\), \(\theta=\frac{\pi}{6},\frac{11\pi}{6}\).
When \(\cos\theta=-\frac{\sqrt{3}}{2}\), \(\theta=\frac{5\pi}{6}+2k\pi\) or \(\theta=\frac{7\pi}{6}+2k\pi\) (\(k\in\mathbb{Z}\)).
In the interval \(0\leq\theta < 2\pi\), when \(k = 0\), \(\theta=\frac{5\pi}{6},\frac{7\pi}{6}\).

Answer:

\(\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\)