SOLVING TRIG INEQUALITIES – UPDATED
(By Nghi H Nguyen, Updated Nov 23, 2020)
GENERALITIES ON TRIG INEQUALITIES
A trig inequality is an inequality in the form F(x) ≤ 0, or F(x) ≥ 0 that contains one or a few trig
functions of the arc AM = x. Solving for x means finding all values of the arc x whose trig
functions make the inequality true.
All these values of arc x (expressed in radians or degrees) constitute the solution set of the trig
inequality.
Solution set of a trig inequality are expressed in the form of intervals, or arc lengths
Example of trig inequalities:
sin x < 1/2
sin x + sin 2x > sin 3x
sin x + cos x ≤ sin 3x
cos² x – sin x ≥ -1
tan x – cot x > 1
(2cos x – 1)(2cos x + 1) < 0
Example of intervals (similar to the Number Line method)
Open intervals:
(0, Ꙥ/3)
(Ꙥ/6, 5Ꙥ/6)
(Ꙥ/2, 3Ꙥ/2)
Half closed intervals:
(-∞, Ꙥ/6]
[Ꙥ/3, +∞)
[0, +∞)
Closed intervals:
[Ꙥ/3, 3Ꙥ/2]
[20⁰, 90⁰]
[45⁰, 120⁰]
THE TRIG UNIT CIRCLE AND TRIG FUNCTIONS
It is a circle with radius R = 1, and with an origin O. The unit circle defines the main trig
functions of the variable arc x that varies and rotates counterclockwise on it.
On the unit circle, the value of the arc x is exactly equal to the corresponding angle x. It is
more convenient to use the arc x as the variable, instead of the angle x.
When the variable arc AM = x (in radians or degree) rotates counterclockwise on the unit
circle:
- The horizontal axis OAx defines the function f(x) = cos x. When the arc x varies from 0 to
2Ꙥ, the function f(x) = cos x varies from 1 to -1, then, back to 1.
- The vertical axis OBy defines the function f(x) = sin x. When the arc x varies from 0 to 2Ꙥ,
the function f(x) = sin x varies from 0 to 1, then to -1, then back to 0.
- The vertical axis AT defines the function f(x) = tan x. When x varies from -Ꙥ/2 to Ꙥ/2, the
function f(x) varies from - ∞ to +∞.
- The horizontal axis BU defines the function f(x) = cot x. When x varies from 0 to Ꙥ, the
function f(x) = cot x varies from +∞ to -∞.
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Figure 1
CONCEPT OF SOLVING TRIG INEQUALITIES.
To solve a trig inequality, transform it into one or many basic inequalities. Solving trig
inequalities finally results in solving trig basic inequalities, or similar
There are 4 basic trig inequalities:
F(x) = sin x ≥ a (or ≤ a)
F(x) = tan x ≤ a (or ≥ a)
F(x) = cos x ≤ a (or ≥ a)
F(x) = cot x ≤ a (or ≥ a)
SOLVING BASIC INEQUALITIES
The unit circle and the 4 axis are used as proof to solve basic trig inequalities, or similar.
Example 1. Solve
cos x > 1/2
Solution. Standard form: F(x) = cos x – 1.2 > 0
First, solve F(x) = 0. On the cos axis, take OM = 1/2. See Figure 2
There are 2 solution arcs AM1 and AM2 that have the same cos value (1/2)
The Table of Special Arcs (in any trig book) give → cos (Ꙥ/3) = cos (-Ꙥ/3) = 1/2.
The 2 end points are (Ꙥ/3) and (-Ꙥ/3).
The solution set of the trig inequality F(x) > 0 is the arc length, or open interval: (-Ꙥ/3, Ꙥ/3)
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Example 2. Solve
sin x ≤ 1/2
Solution.
Standard form: F(x) = sin x – 1/2 ≤ 0.
First, solve F(x) = 0 → sin x = 1/2
Table of special arcs and unit circle gives x = Ꙥ/6. The unit circle gives another arc x = 5Ꙥ/6
By considering the unit circle, we see that F(x) = sin x – 1/2 < 0, when the arc x is inside the
arc length (5Ꙥ/6 , Ꙥ/6 + 2Ꙥ). See Figure 3.
Consequently, the solution set of F(x) = sin x – 1/2 ≤ 0 is the closed interval [5Ꙥ/6, 13Ꙥ/6].
The 2 end points are included in the solution set.
Example 3. Solve
tan x < 3/4.
Solution. Standard form: F(x) = tan x – 3/4 < 0
First, solve F(x) = 0. On the tangent axis, AT = 0.75. Calculator give arc AM1 = 36⁰87 and arc
AM2 = 180⁰ + 36⁰87 = 216⁰87.
By considering the unit circle, we see that F(x) = tan x - 3/4 < 0 when the arc x varies inside
the arc length (- 90⁰, 36⁰87) and the arc length (90⁰, 216⁰87). Color them green.
Note that the arc (-90⁰) is co-terminal to arc (270⁰).
Finally, the solution set of the trig inequality F(x) = tan x – 3/4 < 0 for (0, 360⁰) is the two
open intervals:
(90⁰, 216⁰87) and (270⁰, 396⁰87)
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Figure 4
METHODS TO SOLVE TRIG INEQUALITIES.
So far, there are 2 main methods to solve complex trig inequalities:
1. The algebraic method by creating a Sign Chart (sign table)
2. The innovative method by using the number unit circle, recently introduced by the author of
this article.
1. THE ALGEBRAIC METHOD USING THE SIGN CHART
There are in general 4 steps in solving a complex trig inequalities by using the sign chart.
STEP 1. Transform the inequality into standard form.
Example. The inequality (cos 2x < 2 + 3sin x) will be transformed into standard form
F(x) = cos 2x – 3sin x – 2 < 0
Example. The inequality (sin x + sin 2x > sin 3x) will be transformed into standard form:
F(x) = sin x + sin 2x - sin 3x > 0
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STEP 2. Find the common period
The common period of the complex trig inequality is the least common multiple of all the
periods presented in the inequality.
Examples: The common period of
F(x) = sin x + cos 2x + sin 3x is: 2Ꙥ
F(x) = tan x + 2cot x is: Ꙥ
F(x) = sin 2x + 3cos 2x + sin (x/2) is: 4Ꙥ
STEP 3. Transform the given trig inequality F(x) into a product of one or many simple trig
functions.
a. If F(x) contains only one trig function, there is no need to perform transformation. Solve it as
a basic trig inequality, or similar.
Example.
F(x) = tan (2x – Ꙥ/3) > √2
F(x) = sin (x – 135⁰) < √3/2
b. If the complex inequality F(x) contains 2 or more trig functions, transform F(x) into one or
many simple trig functions in the form:
F(x) = f(x).g(x) ≤ 0 (or ≥ 0)
or
F(x) = f(x).g(x).h(x) ≥0 (or ≤ 0)
To know how to transform F(x) into two or 3 simple trig functions, please read the article titled
“Solving trig equations – concept and method” by the same author.
Examples of complex trig equations:
F(x) = sin x + sin 3x > 0
F(x) = sin (x/2).cos (x – Ꙥ/3) > 0
F(x) = cos 2x + sin 2x < cos x
F(x) = tan 2x.tan x < 1
STEP 4. Creating the Sign Chart
After transformation, we get F(x) = f(x).g(x) < 0 (or > 0)
First step, solve f(x) = 0 to find the 2 end points that divides the unit circle into 2 arc lengths.
Plot the 2 end points in a sign chart with arc x varies inside the common period.
Find the sign status (+ or -) of the trig function f(x) inside the 2 arc lengths.
Do the same thing for the trig function g(x).
Create the sign chart that combines the 2 sign charts of f(x) and g(x).
The resulting sign of F(x) will show the combined solution set of the trig inequality.
Example 4. Solve sin x + sin 2x < - sin 3x
Solution.
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First step → Standard form: F(x) = sin x + sin 2 x + sin 3 x < 0
Second step. Common period → 2Ꙥ
Third step. Use trig identity (sin a + sin b) to transform equation F(x)
F(x) = sin x + sin 3x + sin 2x = 2sin 2x.cos x + sin 2x = sin 2x(2cos x + 1) < 0.
F(x) = f(x).g(x) < 0
1. Next, solve f(x) = sin 2x = 0 → there are 3 end points.
2x = 0 + 2kꙤ and
2x = Ꙥ + 2kꙤ
2x = 2Ꙥ + 2kꙤ
x=0
x = Ꙥ/2
x=Ꙥ
For k = 1 , there is one more end point: x = Ꙥ/2 + Ꙥ = 3Ꙥ/2.
Plot these 4 end points on the sign chart, and find the sign status of f(x) = sin 2x when x varies
inside these 4 arc lengths.
f(x) = sin 2x > 0 when sin x varies inside arc length; (0, Ꙥ/2).
Figure the sign status (+ or -) of f(x) for other arc lengths.
Sign chart of f(x):
x |0
Ꙥ/2
Ꙥ
3Ꙥ/2
2Ꙥ
---------------------------------------------------------------------------------------------------------f(x) |
+
0
0
+
0
0
2. Next solve g(x) = 2cos x + 1 = 0 → cos x = - 1/2 → There are 2 end points at: x = 2Ꙥ/3, and
x = 4Ꙥ/3.
g(x) < 0 inside interval (2Ꙥ/3, 4Ꙥ/3). Set up the sign chart of g(x)
x
|0
2Ꙥ/3
Ꙥ
4Ꙥ/3 3Ꙥ/2
2Ꙥ
--------------------------------------------------------------------------------------------------------g(x) |
+
0
0
+
Finally, create the combined sign chart of F(x) = f(x).g(x)
x |0
Ꙥ/2 2Ꙥ/3 Ꙥ
4Ꙥ/3
3Ꙥ/2
2Ꙥ
-------------------------------------------------------------------------------------------------------f(x) |0
+
0 - | - 0 +
| +
0
0
-------------------------------------------------------------------------------------------------------g(x) |
+
| + 0 - | 0 +
|
+
-------------------------------------------------------------------------------------------------------F(x) |
| yes |
| yes |
|
yes
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The resulting sign of F(x) shows that the combined solution set of the original inequality are
the 3 open intervals:
(Ꙥ/2, 2Ꙥ/3) and (Ꙥ, 4Ꙥ/3) and (3Ꙥ/2, 2Ꙥ)
METHOD 2. THE INNOVATIVE METHOD USING THE NUMBER UNIT CIRCLE
This innovative method uses the number unit circle to visually solve trig inequalities. It
proceeds exactly the same as the number line check point method. The difference is, instead of
a number line, it uses the unit circle that is numbered (in radians or degrees) depending on the
common period. In general, the 2 methods - using sign chart or using number unit circle perform with the same solving process. However the unit circle method is more visual and it
shows the periodic characteristics of the trig functions.
A. Solving basic trig inequalities by the number unit circle.
This method finds the solution set of the trig inequality by using the numbered unit circle. It
proceeds exactly like the number line check point method.
Example 5. Solve
sin x < - 1/2
Solution. Standard form: F(x) = sin x + 1/2 < 0
First step, solve F(x) = sin x + 1/2 = 0 → sin x = - 1/2
Table of special arc gives arc: x = - Ꙥ/6, or x = 11Ꙥ/6 (co-terminal)
The unit circle gives another arc x = 7Ꙥ/6 that has the same sin value (- 1/2)
Figure 4
Plot the 2 points (7Ꙥ/6) and (11Ꙥ/6) on the unit circle. They divides the circle into 2 arc
lengths. Select the point (3Ꙥ/2) as check point. We get F(x) = sin (3Ꙥ/2) - 1/2 = - 1 – 1/2 < 0.
It is true. Therefor, the solution set of F(x) < 0 is the open interval (7Ꙥ/6, 11Ꙥ/6)
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Select the point (3Ꙥ/2) as check point. We get F(x) = sin (3Ꙥ/2) - 1/2 = - 1 – 1/2 < 0.
It is true. Therefor, the solution set of F(x) < 0 is the open interval (7Ꙥ/6, 11Ꙥ/6)
B. Solving complex trig inequalities by the number unit circle
Example 6. Solve: sin x + sin 2x + sin 3x < cos x + cos 2x + cos 3x
(0 < x < 2Ꙥ)
Solution. Use Sum to Product Identity (sin a + sin b) and (cos a + cos b) to transform 2 sides.
sin 2x(2cos x + 1) < cos 2x (2cos x + 1)
F(x) = f(x).g(x) = (2cos x + 1)(sin 2x – cos 2x) < 0
Common period : (2Ꙥ)
1. Solve f(x) = (2cos x + 1) = 0 → cos x = -1/2→ Two end points at: x = 2Ꙥ/3 and x = 4Ꙥ/3
Use check point x = Ꙥ → we get: f(x) = F(Ꙥ) = 0 < 1.
Then, f(x) < 0 inside the interval (2Ꙥ/3, 4Ꙥ/3) (blue). The arc length for f(x) > 0 is colored red
2. Solve. g(x) = sin 2x – cos 2x = 0 → Divide by cos 2x (condition cos 2x ≠ 0)
We get: tan 2x - 1 = 0 → tan 2x = 1 -->2 solutions : 2x = Ꙥ/4 + kꙤ, and 2x = 5Ꙥ/4 + kꙤ
2x = Ꙥ/4 →that gives: x = Ꙥ/8 + kꙤ/2
2x = 5Ꙥ/4 → that gives: x = 5Ꙥ/8 + kꙤ/2
For k = 0; k = 1; and k = 2 → There are 4 end points at: Ꙥ/8, 5Ꙥ/8, 9Ꙥ/8, and 13Ꙥ/8
Use check point x = Ꙥ/2. We get: g(Ꙥ/2) = sin Ꙥ – cos Ꙥ = 0 - (-1) = 1 -- > 0. Then g(x) > 0
inside the interval (Ꙥ/8, 5Ꙥ/8). Color this arc length red.
Color the 3 other arc lengths, following the rule of end points.
Figure 5
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By superimposing, we see that the solution set are the 3 open intervals:
(5Ꙥ/8, 2Ꙥ/3) where f(x) > 0 and g(x) < 0, and
(9Ꙥ/8, 4Ꙥ/3)
where f(x) < 0, and g(x) > 0 and
(13Ꙥ/8, Ꙥ/8 + 2Ꙥ = 17Ꙥ/8) where f(x) > 0 and g(x) < 0
NOTE.
The unit circle can be numbered depending on the common period of the trig inequality.
For examples: From 0 to Ꙥ, or From 0 to 180⁰, or From 0 to 480⁰.
Example 7. Solve
Solution.
tan x. tan 2x < 0
F(x) = f(x).g(x) = tan x. tan 2x < 0
Common period: Ꙥ
f(x) = tan x = 0 → two end points x = 0, and x = Ꙥ.
There is discontinuation at x = Ꙥ/2
f(x) > 0 when x varies inside interval (0, Ꙥ/2). Color it red, and the rest blue.
Figure 6
g(x) = tan 2x = 0 → 2 end points at x = 0, and x = Ꙥ
There are discontinuations at: 2x = Ꙥ/2 → x = Ꙥ/4, and 2x = 3Ꙥ/2 → x = 3Ꙥ/4
g(x) > 0 when x varies inside the interval (0, Ꙥ/4). Color it red and color the other intervals.
By superimposing, we see that the combined solution set are the open intervals:
(Ꙥ/4, Ꙥ/2) and (Ꙥ/2, 3Ꙥ/4)
Check.
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Inequality:
tan x.tan 2x < 0
x = Ꙥ/3 → tan (Ꙥ/3).tan (2Ꙥ/3) → (+).(-) < 0. Proved
x = (2Ꙥ/3) → tan (2Ꙥ/3).tan (4Ꙥ/3) → (-).(+) < 0. Proved
x = Ꙥ/6 → tan (Ꙥ/6).tan (2Ꙥ/6) = tan (Ꙥ/6).tan (Ꙥ/3) → (+).(+). Not true
x = 5Ꙥ/6 → tan (5Ꙥ/6).tan (10Ꙥ/6) = tan (5Ꙥ/6).tan (5Ꙥ/3) → (-).(-). Not true
Remark 1
Students may use the triple number unit circle to solve complex trig inequalities, type F(x) that
can be transformed into 3 simple trig functions: F(x) = f(x).g(x).h(x) < 0 (or > 0)
Remark 2
On the unit circle the value of the arc x, and the corresponding angle x, are exactly equal. The
French concept to select the arc x as variable, instead of the angle x, makes the whole study of
trigonometry convenient, reasonable, visual, and less absurd.
- The popular expressions such as (arc tan x) and (arc sin Ꙥ/3) makes sense with this concept.
- About the trig functions tan x and cot x, when they go to infinities, the infinity notion is
understandable as two axis lines going parallel one to the other
- There are 2 axis AT and BU that define the 2 functions tan x and cot x. Solving the inequality
(tan x < 1) is easy and simple by using the axis AT and the unit circle. However, solving the
inequality (sin x/cox x < 1) becomes complicated, because it is dealing with two function
variables.
- Solving the inequality tan x < a (or > a), by using the tan function’s graph, takes too much
time in creating the graph.
- The extended answer such as (30⁰ + 720⁰) can be visually understood as an arc length of 30⁰
plus 2 circumferences. On the contrary, the flat angle (30⁰ + 720⁰) looks confused with a flat
area.
- In practice, the rules of end points and arc lengths on the unit circle make the solving process
more convenient than handling the sign chart method.
- Finally, the numbered unit circle provides students with a second method to solve complex
trig inequalities, besides the sign chart method.
(This article was written by Nghi H Nguyen, author of the new Transforming Method to solve
quadratic equations – Updated Nov. 23, 2020)
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