use an identity to solve the following equation on the interval 0,2π).\n\nsin 2x = cos x\n\nselect the…

use an identity to solve the following equation on the interval 0,2π).\n\nsin 2x = cos x\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. x=\n(use integers or fractions for any numbers in the expression. type an exact answer, using π as needed. use a comma to separate answers as needed.)\n\nb. there is no solution.

use an identity to solve the following equation on the interval 0,2π).\n\nsin 2x = cos x\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. x=\n(use integers or fractions for any numbers in the expression. type an exact answer, using π as needed. use a comma to separate answers as needed.)\n\nb. there is no solution.

Answer

Explanation:

Step1: Use double - angle identity

Recall the double - angle identity (\sin2x = 2\sin x\cos x). The given equation (\sin2x=\cos x) becomes (2\sin x\cos x-\cos x = 0).

Step2: Factor out (\cos x)

Factor (\cos x) from the left - hand side: (\cos x(2\sin x - 1)=0).

Step3: Set each factor equal to zero

  • Case 1: (\cos x=0) If (\cos x = 0), then (x=\frac{\pi}{2}+k\pi), (k\in\mathbb{Z}). For (x\in[0,2\pi)), when (k = 0), (x=\frac{\pi}{2}); when (k = 1), (x=\frac{3\pi}{2}).
  • Case 2: (2\sin x-1 = 0) Solve (2\sin x-1 = 0) for (\sin x). We get (\sin x=\frac{1}{2}). If (\sin x=\frac{1}{2}), then (x=\frac{\pi}{6}+2k\pi) or (x=\frac{5\pi}{6}+2k\pi), (k\in\mathbb{Z}). For (x\in[0,2\pi)), when (k = 0), (x=\frac{\pi}{6}) and (x=\frac{5\pi}{6}).

Answer:

(x = \frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2})