Stitz-Zeager_College_Algebra_e-book

# For example n40 e read 40 east of north is a bearing

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Unformatted text preview: it is not in [0, 2π ). We can safely disregard the answers corresponding to k < −1 as well. Next, we move to the family of solutions x = π − 1 arcsin(0.87) + πk for integers k . For 2 2 k = 0, we get x = π − 1 arcsin(0.87). Since 0 < 1 arcsin(0.87) < π , x = π − 1 arcsin(0.87) 2 2 2 4 2 2 lies between π and π so it lies in the speciﬁed range of [0, 2π ). Advancing to k = 1, 4 2 π we get x = π − 1 arcsin(0.87) + π = 32 − 1 arcsin(0.87). Since 0 < 1 arcsin(0.87) < π , 2 2 2 2 4 3π 1 5π π x = 2 − 2 arcsin(0.87) is between 4 and 32 , well within the range [0, 2π ) so we keep it. For π k = 2, we ﬁnd x = π − 1 arcsin(0.87) + 2π = 52 − 1 arcsin(0.87). Since 1 arcsin(0.87) < π , 2 2 2 2 4 5π 1 9π x = 2 − 2 arcsin(0.87) > 4 which is outside the range [0, 2π ). Checking the negative integers, we begin with k = −1 and get x = π − 1 arcsin(0.87) − π = − π − 1 arcsin(0.87). Since 2 2 2 2 π 1 1 2 arcsin(0.87) > 0, x = − 2 − 2 arcsin(0.87) < 0...
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