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Ch. 3 - Trigonometric Identities and Equations
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 127

In Exercises 127–130, solve each equation on the interval [0, 2𝝅) by first rewriting the equation in terms of sines or cosines. csc² x + csc x - 2 = 0

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Start by rewriting the given equation in terms of sine, since \( \csc x = \frac{1}{\sin x} \). The equation is \( \csc^2 x + \csc x - 2 = 0 \), so substitute to get \( \frac{1}{\sin^2 x} + \frac{1}{\sin x} - 2 = 0 \).
To simplify, multiply the entire equation by \( \sin^2 x \) (noting that \( \sin x \neq 0 \) in the domain) to clear the denominators: \( 1 + \sin x - 2 \sin^2 x = 0 \).
Rewrite the equation in standard quadratic form in terms of \( \sin x \): \( -2 \sin^2 x + \sin x + 1 = 0 \). For easier handling, multiply both sides by \( -1 \) to get \( 2 \sin^2 x - \sin x - 1 = 0 \).
Let \( y = \sin x \). Now solve the quadratic equation \( 2y^2 - y - 1 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=2 \), \( b=-1 \), and \( c=-1 \).
After finding the values of \( y \), determine which solutions lie within the range \( [-1, 1] \) since \( \sin x \) must be in this interval. Then, find all \( x \) in \( [0, 2\pi) \) such that \( \sin x = y \).

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