College Trigonometry, 2011a

866 Foundations of Trigonometry
Next, we focus on solving inequalities involving the trigonometric functions. Since these functions
are continuous on their domains, we may use the sign diagram technique we’ve used in the past to
solve the inequalities. 10
Example 10.7.3. Solve the following inequalities on [0, 2π). Express your answers using interval
notation and verify your answers graphically.
1. 2 sin(x) ≤ 1 2. sin(2x) > cos(x) 3. tan(x) ≥ 3
Solution.
1. We begin solving 2 sin(x) ≤ 1 by collecting all of the terms on one side of the equation and
zero on the other to get 2 sin(x) − 1 ≤ 0. Next, we let f(x) =2sin(x) − 1 and note that our
original inequality is equivalent to solving f(x) ≤ 0. We now look to see where, if ever, f is
undefined and where f(x) = 0. Since the domain of f is all real numbers, we can immediately
set about finding the zeros of f. Solving f(x) = 0, we have 2 sin(x) − 1=0orsin(x) = 1 2 .
The solutions here are x = π 6 +2πk and x = 5π 6
+2πk for integers k. Since we are restricting
our attention to [0, 2π), only x = π 6 and x = 5π 6
are of concern to us. Next, we choose test
values in [0, 2π) other than the zeros and determine if f is positive or negative there. For
x =0wehavef(0) = −1, for x = π 2
we get f ( )
π
2 = 1 and for x = π we get f(π) =−1.
Since our original inequality is equivalent to f(x) ≤ 0, we are looking for where the function
is negative (−) or 0, and we get the intervals [ 0, π ] [
6 ∪ 5π
6 , 2π) . We can confirm our answer
graphically by seeing where the graph of y =2sin(x) crosses or is below the graph of y =1.
(−)
0 (+)
0 (−)
0
π
6
5π
6
2π
y =2sin(x) andy =1
2. We first rewrite sin(2x) > cos(x) assin(2x) − cos(x) > 0andletf(x) =sin(2x) − cos(x).
Our original inequality is thus equivalent to f(x) > 0. The domain of f is all real numbers,
so we can advance to finding the zeros of f. Setting f(x) = 0 yields sin(2x) − cos(x) =0,
which, by way of the double angle identity for sine, becomes 2 sin(x) cos(x) − cos(x) =0or
cos(x)(2 sin(x)−1) = 0. From cos(x) = 0, we get x = π 2 +πk for integers k of which only x = π 2
and x = 3π 2
lie in [0, 2π). For 2 sin(x) − 1 = 0, we get sin(x) = 1 2 which gives x = π 6
+2πk or
x = 5π 6 +2πk for integers k. Of those, only x = π 6 and x = 5π 6
lie in [0, 2π). Next, we choose
10 See page 214, Example 3.1.5, page 321, page 399, Example 6.3.2 and Example 6.4.2 for discussion of this technique.