Questions: Solve the equation on the interval (0 leq theta<2 pi). [ 8 sin ^2 theta-17 sin theta+9=0 ] What is the solution in the interval (0 leq theta<2 pi) ? Select the correct choice and fill in any answer boxes in your choice below. A. The solution set is (square) (Simplify your answer. Type an exact answer, using (pi) 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.

Solution Steps

To solve the trigonometric equation \(8 \sin^2 \theta - 17 \sin \theta + 9 = 0\) on the interval \(0 \leq \theta < 2\pi\), we can treat it as a quadratic equation in terms of \(\sin \theta\). First, solve for \(\sin \theta\) using the quadratic formula. Then, find the corresponding \(\theta\) values within the given interval by taking the inverse sine of the solutions and considering the periodic nature of the sine function.

We start with the equation: \[ 8 \sin^2 \theta - 17 \sin \theta + 9 = 0 \] This is a quadratic equation in terms of \( \sin \theta \).

The discriminant \( D \) is calculated as follows: \[ D = b^2 - 4ac = (-17)^2 - 4 \cdot 8 \cdot 9 = 289 - 288 = 1 \] Since the discriminant is positive, there are two distinct real solutions for \( \sin \theta \).

Using the quadratic formula: \[ \sin \theta = \frac{-b \pm \sqrt{D}}{2a} \] Substituting the values: \[ \sin \theta = \frac{17 \pm \sqrt{1}}{16} = \frac{17 \pm 1}{16} \] This gives us two solutions: \[ \sin \theta_1 = \frac{18}{16} = \frac{9}{8} \quad (\text{not valid since } \sin \theta \leq 1) \] \[ \sin \theta_2 = \frac{16}{16} = 1 \]

The valid solution is \( \sin \theta = 1 \). The angle \( \theta \) that satisfies this in the interval \( 0 \leq \theta < 2\pi \) is: \[ \theta = \frac{\pi}{2} \]

Final Answer