sin(2θ)=\\frac{1}{2}\nwhat are the solutions to sin(2θ)=\\frac{1}{2} in the interval 0,2π)? select the…

sin(2θ)=\\frac{1}{2}\nwhat are the solutions to sin(2θ)=\\frac{1}{2} in the interval 0,2π)? select the correct\nchoice and fill in any answer boxes in your choice below.\na. θ=\n(simplify your answer. type an exact answer, using π as needed. type\nyour answer in radians. use integers or fractions for any numbers in the\nexpression. use a comma to separate answers as needed.)\nb. there is no solution.

sin(2θ)=\\frac{1}{2}\nwhat are the solutions to sin(2θ)=\\frac{1}{2} in the interval 0,2π)? select the correct\nchoice and fill in any answer boxes in your choice below.\na. θ=\n(simplify your answer. type an exact answer, using π as needed. type\nyour answer in radians. use integers or fractions for any numbers in the\nexpression. use a comma to separate answers as needed.)\nb. there is no solution.

Answer

Explanation:

Step1: Find the 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}=2k\pi+\frac{5\pi}{6}), (k\in\mathbb{Z}). Since (x = 2\theta), we have (2\theta=2k\pi+\frac{\pi}{6}) or (2\theta=2k\pi+\frac{5\pi}{6}).

Step2: Solve for (\theta)

For (2\theta=2k\pi+\frac{\pi}{6}), we get (\theta=k\pi+\frac{\pi}{12}). For (2\theta=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)

Answer:

(\theta=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12})