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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 15

In Exercises 12–18, solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places. 2 sin² x + cos x = 1

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Start by rewriting the given equation: \(2 \sin^{2} x + \cos x = 1\).
Use the Pythagorean identity \(\sin^{2} x = 1 - \cos^{2} x\) to express the equation entirely in terms of \(\cos x\). Substitute to get: \(2(1 - \cos^{2} x) + \cos x = 1\).
Simplify the equation: \(2 - 2 \cos^{2} x + \cos x = 1\). Then rearrange to form a quadratic equation in \(\cos x\): \(-2 \cos^{2} x + \cos x + 1 = 0\).
Multiply the entire equation by \(-1\) to make the quadratic standard: \(2 \cos^{2} x - \cos x - 1 = 0\). Now solve this quadratic equation for \(\cos x\) using the quadratic formula \(\cos x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) where \(a=2\), \(b=-1\), and \(c=-1\).
After finding the values of \(\cos x\), determine the corresponding values of \(x\) in the interval \([0, 2\pi)\) by using the inverse cosine function and considering the cosine sign in different quadrants.

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The problem restricts solutions to the interval [0, 2π), meaning all solutions must be found within one full rotation of the unit circle. Understanding how to interpret this interval and identify all valid solutions within it is crucial for providing a complete answer.
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