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Unformatted text preview: = tan(arctan(117)) − 3 = 114. We choose our
last test value to be x = 74 and ﬁnd f 74 = −4. Since we want f (x) ≥ 0, we see that our
answer is arctan(3), π ∪ arctan(3) + π, 32 . Using the graphs of y = tan(x) and y = 3, we
can verify when the graph of the former is above (or meets) the graph of the latter.
10 See page 247 for a discussion of the non-standard character known as the interrobang.
We could have chosen any value arctan(t) where t > 3.
. . . by adding π through the inequality . . .
11 10.7 Trigonometric Equations and Inequalities (−) 0 (+)
0 (−) 0 (+)
2 arctan(3) (arctan(3) + π ) 739 (−)
y = tan(x) and y = 3 We close this section with an example that puts solving equations and inequalities to good use –
ﬁnding domains of functions.
Example 10.7.4. Express the domain of the following functions using extended interval notation.13
1. f (x) = csc 2x + π
3 2. f (x) = sin(x)
2 cos(x) − 1 3. f (x) = 1 − cot(x) Solution.
1. To ﬁnd the domain of f (x) = csc 2x + π
3 , we rewrite f in terms of sine as f (x) = 1
sin(2x+ π )
3 Since the sine function is deﬁned everywhere, our only concern comes from zeros in the denominator. Solving sin 2x + π = 0, we get...
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