Stitz-Zeager_College_Algebra_e-book

# 5 b 732 541 a 117 l 6 a 57 b 100 c

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Unformatted text preview: sin 2 π − 1 arcsin(0.87) + πk = sin (π − arcsin(0.87) + 2πk ) = sin (π − arcsin(0.87)) = 2 2 0.87. We now need to determine which of our solutions lie in [0, 2π ). Starting with the family of solutions x = 1 arcsin(0.87)+ πk , we ﬁnd k = 0 gives x = 1 arcsin(0.87). Since 0.87 &gt; 0, we 2 2 1 know that 0 &lt; arcsin(0.87) &lt; π . Dividing through by 2 gives 0 &lt; 2 arcsin(0.87) &lt; π . Hence 2 4 1 x = 2 arcsin(0.87) lies in the interval [0, 2π ). Next, we let k = 1 and get x = 1 arcsin(0.87)+ π . 2 1 π Since 0 &lt; 2 arcsin(0.87) &lt; π , x = 1 arcsin(0.87) + π is between π and 54 , so we keep this 4 2 1 1 answer as well. When k = 2, we get x = 2 arcsin(0.87) + 2π . Since 2 arcsin(0.87) &gt; 0, x = 1 arcsin(0.87) + 2π &gt; 2π , so we discard this answer along with all answers corresponding 2 to k &gt; 2. Since k represents an integer, we need to allow negative values of k as well. For k = −1, we get x = 1 arcsin(0.87)−π , and since 1 arcsin(0.87) &lt; π , x = 1 arcsin(0.87)−π &lt; 0, 2 2 4 2 so...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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