solve the equation on the interval 0,2π). sin(2θ)=\\frac{1}{2} what are the solutions to sin(2θ)=\\frac{1}{2}…
solve the equation on the interval 0,2π). sin(2θ)=\\frac{1}{2} what are the solutions to sin(2θ)=\\frac{1}{2} in the interval 0,2π)? select the correct choice and fill in any answer boxes in your choice below. a. θ= (simplify your answer. type an exact answer, using π as needed. type your answer in radians. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.) b. there is no solution.
Answer
Explanation:
Step1: Find general solutions for (2\theta)
We know that if (\sin x = \frac{1}{2}), then (x = 2k\pi+\frac{\pi}{6}) or (x=(2k + 1)\pi-\frac{\pi}{6}), (k\in\mathbb{Z}). Since (x = 2\theta), we have (2\theta=2k\pi+\frac{\pi}{6}) or (2\theta=(2k + 1)\pi-\frac{\pi}{6}), (k\in\mathbb{Z}).
Step2: Solve for (\theta)
For (2\theta=2k\pi+\frac{\pi}{6}), we get (\theta=k\pi+\frac{\pi}{12}). For (2\theta=(2k + 1)\pi-\frac{\pi}{6}=2k\pi+\pi-\frac{\pi}{6}=2k\pi+\frac{5\pi}{6}), we get (\theta=k\pi+\frac{5\pi}{12}).
Step3: Find solutions in the interval ([0,2\pi))
When (k = 0): For (\theta=k\pi+\frac{\pi}{12}), (\theta=\frac{\pi}{12}); for (\theta=k\pi+\frac{5\pi}{12}), (\theta=\frac{5\pi}{12}). When (k = 1): For (\theta=k\pi+\frac{\pi}{12}), (\theta=\pi+\frac{\pi}{12}=\frac{13\pi}{12}); for (\theta=k\pi+\frac{5\pi}{12}), (\theta=\pi+\frac{5\pi}{12}=\frac{17\pi}{12}). When (k = 2), (\theta=2\pi+\frac{\pi}{12}>2\pi) and (\theta=2\pi+\frac{5\pi}{12}>2\pi) (rejected).
Answer:
(\theta=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12})