Stitz-Zeager_College_Algebra_e-book

# To do this we take the ratio of their 6 frequencies

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Unformatted text preview: ce, all solutions here are 1 of the form θ = π − α + 2πk = π − arcsin 3 + 2πk , for integers k . 2. Even though we are told t represents a real number, it we can visualize this problem in terms of angles on the Unit Circle, so at least mentally,11 we cosmetically change the equation to √ We could solve x2 = 4 using square roots as well to get x = ± 4, but, we would simplify the answers to x = ±2. 11 In practice, this is done mentally, or in a classroom setting, verbally. Carl’s penchant for pedantry wins out here. 10 718 Foundations of Trigonometry tan(θ) = −2. Tangent is negative in two places: in Quadrant II and Quadrant IV. If we proceed as above using a reference angle approach, then the reference angle α must satisfy 0 < α < π and tan(α) = 2. Such an angle is α = arctan(2) radians. A Quadrant II 2 angle with reference angle α is π − α. Hence, the Quadrant II solutions to the equation are θ = π − α + 2πk = π − arctan(2) + 2πk for integers k . A Quadrant IV angle with reference angle α is 2π...
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