solve the equation on the interval 0, 2π).\n2 sin²θ - 3 sinθ + 1 = 0\nwhat is the solution in the interval…

solve the equation on the interval 0, 2π).\n2 sin²θ - 3 sinθ + 1 = 0\nwhat is the solution in the interval 0, 2π)? select the correct choice 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 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).\n2 sin²θ - 3 sinθ + 1 = 0\nwhat is the solution in the interval 0, 2π)? select the correct choice 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 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: Factor the quadratic equation

Let (x = \sin\theta). The equation (2\sin^{2}\theta - 3\sin\theta+1 = 0) becomes (2x^{2}-3x + 1=0). Factor (2x^{2}-3x + 1): [ \begin{align*} 2x^{2}-3x + 1&=2x^{2}-2x-x + 1\ &=2x(x - 1)-(x - 1)\ &=(2x - 1)(x - 1) \end{align*} ] So, ((2\sin\theta-1)(\sin\theta - 1)=0)

Step2: Solve for (\sin\theta)

Set each factor equal to zero:

  • If (2\sin\theta-1 = 0), then (\sin\theta=\frac{1}{2})
  • If (\sin\theta - 1=0), then (\sin\theta=1)

Step3: Find (\theta) in the interval ([0,2\pi))

  • When (\sin\theta=\frac{1}{2}), (\theta=\frac{\pi}{6}) or (\theta=\frac{5\pi}{6}) (since (\sin\theta=\frac{1}{2}) when (\theta = 30^{\circ}=\frac{\pi}{6}) and (\theta = 150^{\circ}=\frac{5\pi}{6}) in the unit - circle)
  • When (\sin\theta = 1), (\theta=\frac{\pi}{2}) (since (\sin\theta = 1) when (\theta = 90^{\circ}=\frac{\pi}{2}) in the unit - circle)

Answer:

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