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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
Capitolo 7, Problema 6.3.53

Solve each equation over the interval [0, 2π). Write solutions as exact values or to four decimal places, as appropriate
tan 2x + sec 2x = 3

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Start with the given equation: \(\tan 2x + \sec 2x = 3\).
Recall the definitions: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\). Substitute these into the equation to get \(\frac{\sin 2x}{\cos 2x} + \frac{1}{\cos 2x} = 3\).
Combine the terms over the common denominator \(\cos 2x\): \(\frac{\sin 2x + 1}{\cos 2x} = 3\).
Multiply both sides by \(\cos 2x\) to clear the denominator: \(\sin 2x + 1 = 3 \cos 2x\).
Rewrite the equation as \(\sin 2x - 3 \cos 2x = -1\) and use the identity for a linear combination of sine and cosine: \(a \sin \theta + b \cos \theta = R \sin(\theta + \alpha)\), where \(R = \sqrt{a^2 + b^2}\) and \(\alpha = \arctan(\frac{b}{a})\). Apply this to express the left side as a single sine function and then solve for \(x\) over \([0, 2\pi)\).

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