Trigonometry Inequality
Note
the graph of the sine function f(x) = sin x0 in the interval 0 ≤ x ≤ 360 and a
line y = 1/2 below. (See Figure 1).
On the
graph, it appears that the curve with a thick line is a curve that is located
above the line y = 1/2, which is located at the interval 30 < x <150.
he
graph illustrates the values of sin x0 > 1/2.
The
discussion above is an illustration of a completion of trigonometric inequalities.
Thus, a way to solve trigonometric inequalities is to sketch the graph of the
trigonometric function. whereas to determine the interval boundaries, we can
use the trigonometric equation that we discussed in previous trigonometric
articles.
Example 1.
Determine the value of x that satisfies
cos x0 < ½ √3, at intervals 0 ≤ x ≤
360.
Answer:
The
inequality limit value:
cos x0 = ½ √3
x = 30 and x = 330
taking
into account the chart sketch below, the x value that meets the inequality lies
at the interval 30 ≤ x ≤
330.
Example 2.
Determine the value of x that satisfies
tan x0 ≥ - √3, at intervals 0 ≤ x ≤
360.
Answer:
The inequality limit value:
tan x0 = √3
x = 120 and x = 300
with regard to the chart sketch below
and the tangent function not defined at x = 90 and x = 270, then the x value
that meets the inequality lies at the interval.
0 ≤ x ≤ 90, 120 ≤ x ≤ 270, atau 330 ≤ x ≤
360.
Example 3.
Determine the value of x that satisfies
sin 2x0 < - ½, at intervals 0 ≤ x ≤ 360.
Answer;
Sin 2x0 = - ½
2x = 210 + n . 360 or 2x
= - 30 + n . 360
x = 105 + n . 180 or x =
- 15 + n . 180
x = 105, 165, 285, 345
Taking
into account the sketch of the graph above, the x value that meets the
inequality lies in the interval.
105 < x < 165 and 285
< x < 345.
To
solve more complex questions, graph methods are not easy to use. A simpler way
is by the number method, as discussed below.
Example 4:
Determine the x value that satisfies sin
2x0 < - ½, for 0 ≤ x ≤ 360.
Answer:
The solution method is as follows:
Change the inequality so that the right
segment is zero,
Sin 2x0 + ½ < 0
Make a zero value of the sin function 2x0
+ ½, and the value of each is:
105, 165, 285 and 345 (see figure 4
above)
Make a
number line that contains the zero value of the function at 0 ≤ x ≤ 360
intervals
a. Determine the sign of the function at each interval, in the following way.
b. Determine the sign of the function at x = 0 and x = 360, in this case positive
Determine the sign of the value of the function on one of the intervals by substituting
one of the x values located at the interval in question.
Example:
For x = 30, sin 600 + ½ = ½ √3 + ½. So that the
function mark in the interval 0 < x <105 is positive. And so on so that a
number line is marked below.
The values that meet the inequality are located at the
interval marked negative (<0),
i.e. 105 < x < 1165 or 285
< x < 345
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