solve the equation on the interval 0,2π).\n\n\\( \\sin ( 2 \\theta ) = \\frac { 1 } { 2 } \\)\n\nwhat are…

solve the equation on the interval 0,2π).\n\n\\( \\sin ( 2 \\theta ) = \\frac { 1 } { 2 } \\)\n\nwhat are the solutions to \\( \\sin ( 2 \\theta ) = \\frac { 1 } { 2 } \\) in the interval 0,2π)? select the correct choice and fill in any answer boxes in your choice below.\n\na. \\( \\theta = \\) (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.)\nb. there is no solution.

solve the equation on the interval 0,2π).\n\n\\( \\sin ( 2 \\theta ) = \\frac { 1 } { 2 } \\)\n\nwhat are the solutions to \\( \\sin ( 2 \\theta ) = \\frac { 1 } { 2 } \\) in the interval 0,2π)? select the correct choice and fill in any answer boxes in your choice below.\n\na. \\( \\theta = \\) (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.)\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}), where (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}), then (\theta=k\pi+\frac{\pi}{12}). For (2\theta=2k\pi+\frac{5\pi}{6}), then (\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): For (\theta=k\pi+\frac{\pi}{12}), (\theta=2\pi+\frac{\pi}{12}>2\pi); for (\theta=k\pi+\frac{5\pi}{12}), (\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})