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Stitz-Zeager_College_Algebra_e-book

# Theorem 114 suppose a b and c are the angle side

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Unformatted text preview: tan(−3) > −π , x = 2 arctan(−3) + 4π > 3π > 2π so it is outside the range [0, 2π ). Hence we discard it, and all of the solutions corresponding to k > 2 as well. Graphically, we see y = tan x and y = −3 intersect only once on [0, 2π ), and the calculator gives the 2 same decimal approximation for both x = 2 arctan(−3) + 2π and the x-coordinate of the lone intersection point, which is x ≈ 3.7851. 6. As 0.87 isn’t one of the ‘common’ values for sine, we’ll need to use the arcsine function to solve sin(2x) = 0.87. There are two values in [0, 2π ) with a sine of 0.87: arcsin(0.87) and π − arcsin(0.87). Since the period of sine is 2π , we get 2x = arcsin(0.87) + 2πk or 1 2x = π − arcsin(0.87) + 2πk for integers k . Solving for x, we ﬁnd x = 2 arcsin(0.87) + πk or x = π − 1 arcsin(0.87) + πk for integers k . To check, we note that for integers 2 2 1 k , sin 2 2 arcsin(0.87) + πk = sin (arcsin(0.87) + 2πk ) = sin (arcsin(0.87)) = 0.87 and...
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