Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth, as appropriate.
sin² θ ― 2 sin θ + 3 = 0

Verified step by step guidance

1. Recognize that the equation is a quadratic in terms of \( \sin \theta \). Let \( x = \sin \theta \), so the equation becomes \( x^2 - 2x + 3 = 0 \).
2. Use the quadratic formula to solve for \( x \): \( x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \).
3. Calculate the discriminant \( \Delta = (-2)^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 \). Since the discriminant is negative, there are no real solutions for \( x = \sin \theta \).
4. Recall that \( \sin \theta \) must be a real number between -1 and 1, so no real values of \( \theta \) satisfy the equation in the interval \( [0^\circ, 360^\circ) \).
5. Conclude that the equation has no solutions for \( \theta \) in the given interval because the quadratic in \( \sin \theta \) has no real roots.

Key Concepts

Solving Quadratic Equations in Trigonometric Functions

Many trigonometric equations can be rewritten as quadratic equations by substituting a trigonometric expression, such as sin θ, with a variable. This allows the use of algebraic methods like factoring or the quadratic formula to find possible values of the trigonometric function.

Range and Values of the Sine Function

The sine function outputs values only between -1 and 1. When solving equations like sin² θ - 2 sin θ + 3 = 0, it is important to check if the solutions for sin θ fall within this range, as values outside it are not possible and thus yield no valid angle solutions.

Finding Angles from Sine Values within a Given Interval

Once the sine values are found, the corresponding angles θ must be determined within the specified interval [0°, 360°). This involves using the inverse sine function and considering the sine function’s symmetry in the unit circle to find all valid solutions.