Question

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

Explanation:

Step1: Use double - angle formula

Recall the double - angle formula \(\cos(2\theta)=1 - 2\sin^{2}\theta\). Substitute it into the given equation \(\cos(2\theta)+4\sin^{2}\theta = 2\).
We get \(1-2\sin^{2}\theta + 4\sin^{2}\theta=2\).

Step2: Simplify the equation

Combine like terms: \(1 + 2\sin^{2}\theta=2\).
Then, \(2\sin^{2}\theta=2 - 1=1\). So, \(\sin^{2}\theta=\frac{1}{2}\), and \(\sin\theta=\pm\frac{\sqrt{2}}{2}\).

Step3: Solve for \(\theta\)

When \(\sin\theta=\frac{\sqrt{2}}{2}\), \(\theta=\frac{\pi}{4}+2k\pi\) or \(\theta=\frac{3\pi}{4}+2k\pi,k\in\mathbb{Z}\).
When \(\sin\theta=-\frac{\sqrt{2}}{2}\), \(\theta=\frac{5\pi}{4}+2k\pi\) or \(\theta=\frac{7\pi}{4}+2k\pi,k\in\mathbb{Z}\).
Since \(0\leq\theta<2\pi\), when \(k = 0\), the solutions are \(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\).

Answer:

\(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\)