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Ch. 6 - Inverse Circular Functions and Trigonometric Equations
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161, 9780135440780, 9780136881117, 9780135440827, 9780135924891Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoLial 12th EditionCh. 6 - Inverse Circular Functions and Trigonometric EquationsProblema 6.2.43
Capitolo 7, Problema 6.2.43

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

Guida verificata passo dopo passo
1
Recognize that the equation is a quadratic in terms of \( \sin \theta \). Let \( x = \sin \theta \), so the equation becomes \( x^2 - 2x + 3 = 0 \).
Use the quadratic formula to solve for \( x \): \( x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \).
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 \( x = \sin \theta \).
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 \( [0^\circ, 360^\circ) \).
Conclude that the equation has no solutions for \( \theta \) in the given interval because the quadratic in \( \sin \theta \) has no real roots.

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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.
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Solving Quadratic Equations by Completing the Square

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.
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Domain and Range of Function Transformations

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.
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Sine, Cosine, & Tangent of 30°, 45°, & 60°
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