Stitz-Zeager_College_Algebra_e-book

# Rearranging b2 a 2 c2 2ac cos we nd cos a cacb

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Unformatted text preview: + 2πk or 6 9 See page 160, Example 3.1.5, page 247, page 313, Example 6.3.2 and Example 6.4.2 for discussion and examples of this technique. 738 Foundations of Trigonometry x= 5π 6 + 2πk for integers k . Of those, only x = π 6 and x = our test values. For x = 0 we ﬁnd f (0) = −1; when x = for x = 3π 4 we get f 3π 4 √ = −1 + √ 2 2 √ π 4 5π 6 lie in [0, 2π ). Next, we choose we get f π 4 √ = 1− 2 2 = √ 2− 2 2; 2−2 ; when x = π we have f (π ) = 1, and lastly, for √2 −2− 2 ππ . We see f (x) > 0 on π , π ∪ 56 , 32 , so this is 2 62 = π π x = 74 we get f 74 = −1 − 22 = our answer. We can use the calculator to check that the graph of y = sin(2x) is indeed above the graph of y = cos(x) on those intervals. (−) 0 (+) 0 (−) 0 (+) 0 (−) 0 π 6 π 2 5π 6 3π 2 2π y = sin(2x) and y = cos(x) 3. Proceeding as in the last two problems, we rewrite tan(x) ≥ 3 as tan(x) − 3 ≥ 0 and let π f (x) = tan(x) − 3. We note that on [0, 2π ), f is undeﬁned at x = π and 32 , so those 2 values will need the usual disclai...
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