Stitz-Zeager_College_Algebra_e-book

# 38 90 a5 b 12 c 13 d 63 541 629 a 22 b

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Unformatted text preview: Next, we recall that for the square root to be deﬁned, we need 1 − cot(x) ≥ 0. Unlike the inequalities we solved in Example 10.7.3, we are not restricted here to a given interval. For this reason, we employ the same technique as when we solved equations involving cotangent. That is, we solve the inequality on (0, π ) and then add integer multiples of the period, in this case, π . We let g (x) = 1 − cot(x) and set about making a sign diagram for g over the interval (0, π ) to ﬁnd where g (x) > 0. We note that g is undeﬁned for x = πk for integers k , in particular, at the endpoints of our interval x = 0 and x = π . Next, we look for the zeros of g . Solving g (x) = 0, we get cot(x) = 1 or x = π + πk for integers k 4 and only one of these, x = π , lies in (0, π ). Choosing the test values x = π and x = π , we 4 6 2 √ get g π = 1 − 3, and g π = 1. 6 2 (−) 0 (+) π 4 0 We ﬁnd g (x) > 0 on π 4,π π . Adding multiples of the period we get our solution to c...
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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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