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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 6.2.61

Solve each equation (x in radians and θ in degrees) for all exact solutions where appropriate. Round approximate answers in radians to four decimal places and approximate answers in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures.
(2 tan θ) / (3 - tan² θ ) = 1

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1
Start by letting \(t = \tan \theta\) to simplify the equation. The given equation is \(\frac{2 \tan \theta}{3 - \tan^2 \theta} = 1\), which becomes \(\frac{2t}{3 - t^2} = 1\).
Multiply both sides of the equation by the denominator \((3 - t^2)\) to clear the fraction: \(2t = 3 - t^2\).
Rearrange the equation to standard quadratic form: \(t^2 + 2t - 3 = 0\).
Solve the quadratic equation \(t^2 + 2t - 3 = 0\) using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=2\), and \(c=-3\).
Once you find the values of \(t = \tan \theta\), solve for \(\theta\) by taking the arctangent of each solution. Remember to find all solutions within the specified domain, using the periodicity of the tangent function, and express answers in both radians and degrees as required.

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