**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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