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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 124

In Exercises 121–126, solve each equation on the interval [0, 2𝝅). 3 cos² x - sin x = cos² x

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Start by rewriting the given equation: \(3 \cos^{2} x - \sin x = \cos^{2} x\).
Bring all terms to one side to set the equation equal to zero: \(3 \cos^{2} x - \sin x - \cos^{2} x = 0\), which simplifies to \(2 \cos^{2} x - \sin x = 0\).
Use the Pythagorean identity \(\cos^{2} x = 1 - \sin^{2} x\) to express everything in terms of \(\sin x\): substitute to get \(2(1 - \sin^{2} x) - \sin x = 0\).
Expand and simplify the equation: \(2 - 2 \sin^{2} x - \sin x = 0\), then rearrange to form a quadratic in \(\sin x\): \(-2 \sin^{2} x - \sin x + 2 = 0\).
Multiply the entire equation by \(-1\) to make the quadratic standard: \(2 \sin^{2} x + \sin x - 2 = 0\). Now solve this quadratic equation for \(\sin x\) within the interval \([0, 2\pi)\).

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