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Stitz-Zeager_College_Algebra_e-book

# 17 y 5 sin6x 5 sin8x over 0 16 17 y 5 sin6x

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Unformatted text preview: − α, so the Quadrant IV solutions are θ = 2π − α + 2πk = 2π − arctan(2) + 2πk for integers k . As we saw in Section 10.3, these solutions can be combined.12 One way to describe all the solutions is θ = − arctan(2) + πk for integers k . Remembering that we are solving for real numbers t and not angles θ measured in radians, we write our ﬁnal answer as t = − arctan(2) + πk for integers k . y y 1 1 α 1 x α 1 x Alternatively, we can forgo the ‘angle’ approach altogether and we note that tan(t) = −2 only once on its fundamental period − π , π . By deﬁnition, this happens at the value t = 22 arctan(−2). Since the period of tangent is π , we can capture all solutions by adding integer multiples of π and get our solution t = arctan(−2) + πk for integers k . 3. The last equation we are asked to solve, sec(x) = − 5 , poses two immediate problems. First, 3 we are not told whether or not x represents an angle or a real number. We assume the latter, but note that, once again, we will use angles and the Unit Circle to solve the equation re...
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