SOLVING TRIG INEQUALITIES – UPDATED

(By Nghi H Nguyen, Updated August 20, 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 trigfunctions of the arc AM = x. Solving for x means finding all values of the arc x whose trigfunctions make the inequality true.All these values of x (expressed in radians or degrees) constitute the solution set of the triginequality.Solution set of a trig inequality are expressed in the form of intervals, or arc lengthsExample of trig inequalities:

sin x < 1/2 sin x + cos x ≤ sin 3x tan x – cot x > 1sin x + sin 2x > sin 3x cos² x – sin x ≥ -1 (2cos x – 1)(2cos x + 1) < 0

Example of intervals (similar to Number Line)

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 trigfunctions 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 ismore 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 unitcircle:- The horizontal axis OAx defines the function f(x) = cos x. When the arc x varies from 0 to2Ꙥ, 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, thefunction f(x) varies from - ∞ to +∞.

- The horizontal axis BU defines the function f(x) = cot x. When x varies from 0 to Ꙥ, thefunction 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 triginequalities finally results in solving trig basic inequalities, or similar

There are 4 basic trig inequalities:

F(x) = sin x ≥ a (or ≤ a) F(x) = cos x ≤ a (or ≥ a)F(x) = tan 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 > 0First, solve F(x) = 0. On the cos axis, take OM = 1/2. See Figure 2There 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)Page 2

Example 2. Solve sin x ≤ 1/2

Solution. Standard form: F(x) = sin x –1/2 ≤ 0. First, solve F(x) = 0 → sin x = 1/2Table 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 thearc 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 twoopen 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 2 x + sin 3x is: 2ꙤF(x) = tan x + 2cot x is: ꙤF(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 asa basic trig inequality, or similar.

Example. F(x) = sin (x – 135⁰) < √3/2 F(x) = tan (2x – /3) > Ꙥ √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) = cos 2x + sin 2x < cos xF(x) = sin (x/2).cos (x – Ꙥ/3) > 0 F(x) = tan 2x.tan x < 1< 0

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 < 0

Solution.Page 5

First step → Standard form: F(x) = sin x + sin 2 x + sin 3 x < 0Second step. Common period → 2Ꙥ

Third step. Use trig identity (sin a + sin b) to transform equation F(x) = f(x).g(x) < 0

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

Next, solve f(x) sin 2x = 0 → there are 3 end points.2x = 0 + 2kꙤ and 2x = Ꙥ + 2kꙤ 2x = 2Ꙥ + 2kꙤx = 0 x = Ꙥ/2 x = ꙤFor k = 1 , there is one more end point: x = Ꙥ/2 + Ꙥ = 3Ꙥ/2.Plot these 4 end points on the unit circle, and find the sign status of f(x) = sin 2x when x varies inside the 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 intervals.Sign chart of f(x):

x | 0 Ꙥ/2 Ꙥ 3Ꙥ/2 2Ꙥ ----------------------------------------------------------------------------------------------------------f(x) | + 0 - 0 + 0 - 0

Next solve g(x) = 2cos x + 1 = 0 → cos x = - 1/2 → 2 end pointsx = 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. Itproceeds exactly the same as the number line check point method. The difference is, instead ofa number line, it uses the unit circle that is numbered (in radians or degrees) depending on thecommon 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 itshows 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. Itproceeds exactly like the number line check point method.

Example 5. Solve sin x < - 1/2

Solution. Standard form: F(x) = sin x + 1/2 < 0First step, solve F(x) = sin x + 1/2 = 0 → sin x = - 1/2Table 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 4Plot 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)Page 7

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 to transform both sides. Common Period 2Ꙥ

sin 2x(2cos x + 1) < cos 2x (2cos x + 1)F(x) = (2cos x + 1)(sin 2x – cos 2x) < 0F(x) = f(x).g(x) < 0

Solve f(x) = (2cos x + 1) = 0 → cos x = -1/2→ 2 solutions: x = 2Ꙥ/3 and x = 4Ꙥ/3Use check point x = Ꙥ → f(x) = F(Ꙥ) = 0 < 1.Then, f(x) < 0 inside the interval (2Ꙥ/3, 4Ꙥ/3) (blue). The solution set for f(x) > 0 is colored red

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 and 2 end points: 2x = Ꙥ/4 and 2x = 5Ꙥ/4

2x = Ꙥ/4 → x = Ꙥ/8 and 2x = 5Ꙥ/4 → x = 5Ꙥ/8Use check point x = Ꙥ/2. We get g(Ꙥ/2) = sin Ꙥ – cos Ꙥ = 1 – 0 > 0. Then g(x) > 0 inside the interval (Ꙥ/8, 5Ꙥ/8). Draw the solution set of g(x) > 0 (red) and the rest g(x) < 0 (blue)

Figure 5

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By superimposing, we see that the solution set are the 2 open intervals:

(5Ꙥ/8, 2Ꙥ/3) where f(x) > 0 and g(x) < 0, and

(4Ꙥ/3, 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 tan x. tan 2x < 0

Solution. 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. Inequality: tan x.tan 2x < 0

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x = Ꙥ/3 → tan (Ꙥ/3).tan (2Ꙥ/3) → (+).(-) < 0. Provedx = (2Ꙥ/3) → tan (2Ꙥ/3).tan (4Ꙥ/3) → (-).(+) < 0. Provedx = Ꙥ/6 → tan (Ꙥ/6).tan (2Ꙥ/6) = tan (Ꙥ/6).tan (Ꙥ/3) → (+).(+). Not truex = 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.

- The two axis AT (for tan x) and BU (for cot x) make their study simpler and more convenient than the two relations (sin x/cos x) and (cos x/sin x) that are dealing with two function variables.When tan x and cot x go to infinities on their axis, the infinity notion is easily understood as two axis going parallel to each other. On the contrary, the notion of infinity of the relations (sin x/cos x) and (cos x/sin x) is somewhat absurd.Solving the inequality (tan x < 1) is easy and simple by using the axis AT. However, solving theinequality (sin x/cox x < 1) becomes complicated, because it is dealing with two function variables.

- The extended answer such as (30⁰ + 720⁰) can be visually understood as an arc length of 30⁰plus 2 circumferences. On the contrary, the flat angle (30⁰ + 720⁰) looks confused with a flat area.

- Finally, the numbered unit circle provides students with a second method to solve trig inequalities, besides the sign chart method.

(This article was written by Nghi H Nguyen, author of the new Transforming Method to solve quadratic equation – Updated August 20, 2020)

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