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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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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