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.

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.

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.

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.

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