SOLVING TRIG INEQUALITIES – SHOWCASE EXERCISES II
(Authored by Nghi H Nguyen – Updated Mar. 01, 2021)
This article solves a few selected trig inequalities, using Nghi Nguyen method that has
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Exercise 1.
Solve
cos 3x > 0
(period 2Ꙥ)
Solution. Use these trig identities to transform cos 3x:
cos (a + b) = cos a.cos b – sin a.sin b
sin 2a = 2sin.cos a
1 – cos 2a = 2sin² a. We get:
F(x) = cos (x+ 2x) = cos x.cos 2x – sin x.sin 2x > 0
F(x) = cos x.cos 2x – 2sin² x.cos x = cos x(cos 2x – 2sin² x) > 0
F(x) = cos x (cos 2x – 1 + cos 2x) = cos x(2cos 2x – 1) > 0
1. Solve f(x) = cos x = 0. The function f(x) = cos x is positive (> 0) inside the interval
(-Ꙥ/2, Ꙥ/2). Color it red and color the other half circle blue.
2. Solve g(x) = 2cos 2x – 1 = 0. This gives 2 solutions:
2cos 2x = 1 --→ cos 2x =1/2 = cos (± Ꙥ/3) + 2kꙤ--→ x = (± Ꙥ/6) + kꙤ.
There are 4 end points at (Ꙥ/6), (5Ꙥ/6), (7Ꙥ/6) and (11Ꙥ/6), and 4 arc lengths.
Find the sign status of g(x) by selecting point (0) as check point. We get :
g(0) = 2cos 0 – 1 = 2 – 1 = 1 > 0. Therefor, g(x) > 0 inside the interval (11Ꙥ/6, 13Ꙥ/6).
Color it red and color the other arc lengths.
Figure 1
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Figure 2
By superimposing, we see that the solution set of F(x) = cos 3x > 0 are the 3 open
intervals: (Ꙥ/2, 5Ꙥ/6), (7Ꙥ/6, 3Ꙥ/2) and (11Ꙥ/6, 13Ꙥ/6). See Figure 1
Note.
1. We see that the sign chart has 2 separate extremities at x = 0 and x = 2Ꙥ. In this
new method, the 2 extremities joint together and show the periodic character of trig
functions. This is an advantage of the Nghi Nguyen Method.
2. This exercise aims to help students be familiar with the solving practice of complex
trig inequalities by using the number unit circle. There is another method that directly
solves the inequality cos 3x > 0 as a basic trig inequality.
Solve cos 3x = 0. That leads to 2 solutions;
3x = Ꙥ/2 + 2kꙤ. This gives x = Ꙥ/6 + 2kꙤ/3
3x = 3Ꙥ/2 + 2kꙤ. This gives x = Ꙥ/2 + 2kꙤ/3
For k = 0, k = 1, k = 2, there are 6 end points: Ꙥ/6, Ꙥ/2, 5Ꙥ/6, 7Ꙥ/6, 3Ꙥ/2, and 11Ꙥ/6
There are 6 equal arc lengths. The solution set for cos 3x > 0 are the 3 open intervals:
(Ꙥ/2, 5Ꙥ/6) and (7Ꙥ/6, 3Ꙥ/2) and (11Ꙥ/6, 13Ꙥ/6). Same answer as above.
Exercise 2. Solve
sin 3x < 0
Solution. Use these trig identities to transform the inequality.
sin (a + b) = (sin a.cos b + sin b.cos a)
sin 2a = 2sin a.cos a
1 + cos 2a = 2cos² a.
F(x) = sin (x + 2x) = (sin x.cos 2x + sin 2x.cos x) = sin x.cos 2x + 2sin x.cos² x < 0
F(x) = sin x(cos 2x + 2cos² x) = sin x(cos 2x + 2cos² x) > 0
F(x) = sin x(2cos 2x + 1) > 0
1. Solve f(x) = sin x. This function f(x) = sin x is positive (> 0) inside the interval (0, Ꙥ).
Color it red and color the other half blue.
2. Solve g(x) = 2cos 2x + 1 = 0. This gives: cos 2x = - 1/2 ==> 2x = (± 2Ꙥ/3) + 2kꙤ,
and x = (± Ꙥ/3) + kꙤ.
For k = 0 and k = 1, there are 4 end points at: (Ꙥ/3), (2Ꙥ/3), (4Ꙥ/3), and (5Ꙥ/4).
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Find the sign status of g(x) = 2cos2x + 1 by selecting point (0) as check point. We get
g(0) = 2cos (0) + 1 = 3 > 0. Therefore, g(x) > 0 inside this interval (11Ꙥ/6, 13Ꙥ/6).
The solution set for f(x) = sin 3x < 0 are the 3 open intervals:
(Ꙥ/3, 2Ꙥ/3), and (Ꙥ, 4Ꙥ/3), and (5Ꙥ/3, 2Ꙥ). See Figure 2
Note. We can directly solve the inequality (sin 3x < 0) as a basic trig inequality. The
answer will be the same.
Exercise 3. Solve
sin 4x > 0
(period 2Ꙥ)
Solution. Use the trig identity (sin 2a = 2sin a.cos a) to transform.
F(x) = sin 4x = 2sin 2x.cos 2x = 4sin x.cos x.cos 2x > 0
We have the form F(x) = f(x).g(x).h(x) > 0.
1. Solve f(x) = sin x = 0. There are 2 end points at 0 and Ꙥ. The function f(x) = sin x is
positive inside the interval (0, Ꙥ). Color it red and the rest blue.
2. Solve g(x) = cos x = 0. The function g(x) = cos x is positive (> 0) inside the interval
(-Ꙥ/2, Ꙥ/2). Color it red and the rest blue.
3. Solve h(x) = cos 2x = 0. This give 2 solutions:
a. 2x = Ꙥ/2 + 2kꙤ --→ x = Ꙥ/4 + kꙤ
b. 2x = 3Ꙥ/2 + 2kꙤ --→ x = 3Ꙥ/4 + kꙤ
For k = 0 and k = 1, there are 4 end points at: (Ꙥ/4), (3Ꙥ/4), (5Ꙥ/4), and (7Ꙥ/4). There
are 4 equal arc lengths.
Select point (0) as check point. We get h(0) = cos (0) = 1 > 0. Therefore, h(x) is
positive (> 0), inside the interval (- Ꙥ/4, Ꙥ/4). Color it red and color the other arc
lengths.
By superimposing, the combined solution set of F(x) > 0 are the 4 open intervals:
(0, Ꙥ/4) where f(x) > 0, g(x) > 0 and h(x) > 0, and
(Ꙥ/2, 3Ꙥ/4) where f(x) > 0, g(x) < 0 and h(x) < 0, and
(Ꙥ, 5Ꙥ/4) where f(x) < 0, g(x) < 0 and h(x) > 0, and
(3Ꙥ/2, 7Ꙥ/4) where f(x) < 0, g(x) > 0, and h(x) < 0. See Figure 3.
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Figure 3
Exercise 4. Solve
Figure 4
cos 4x > 0
Solution. Use trig identity (1 + cos 2a = 2cos² a) to transform:
F(x) = cos 4x = 2cos² 2x – 1 = (√2cos 2x – 1)(√2cos 2x + 1) > 0.
1. Solve f(x) = √2cos 2x – 1 = 0 ---> cos 2x = √2/2 = cos (± Ꙥ/4) --→ x = (± Ꙥ/8) + kꙤ.
For k = 0, k = 1, there are 4 end points at: (±Ꙥ/8) and (± 9Ꙥ/8), and 4 arc lengths.
Select the check point (x = 0), we get: f(0) = √2 – 1 > 0. Therefore, f(x) > 0 inside the
interval (- Ꙥ/8, Ꙥ/8). Color it red and color the other arc lengths.
2. Solve g(x) = √2cos 2x + 1 = 0 --→ 2x = - √2/2 = cos (± 3Ꙥ/4) + 2kꙤ --→
x = (± 3Ꙥ/8) + kꙤ.
For k = 0, and k = 1, there are 4 end points at (± 3Ꙥ/8) and (± 11Ꙥ/8) or at:
(3Ꙥ/8), (5Ꙥ/8), (7Ꙥ/8), and (11Ꙥ/8).
Select the point (Ꙥ/2) as check point. We get: g(Ꙥ/2) = √2cos Ꙥ + 1 = - √2 + 1 < 0.
Therefore, g(x) < 0 inside the interval (3Ꙥ/8, 5Ꙥ/8). Color it blue and color the 3 other
intervals.
By superimposing, the solution set of F(x) = cos 4x > 0 are the 4 open intervals
(3Ꙥ/8, 5Ꙥ/8) and (7Ꙥ/8, 9Ꙥ/8), and (11Ꙥ/8, 13Ꙥ/8), and (- Ꙥ/8, Ꙥ/8). See Figure 4.
Note. We can directly solve the inequalities sin 4x > 0 and cos 4x > 0 as basic trig
inequalities.
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Exercise 5. Solve
tan x + 2tan² x < cot x + 2
Solution. Call t = tan x, we have
F(t) = t + 2t² – 1/t - 2 < 0
F(t) = t² + 2t³ - 1 - 2t < 0
F(t) = t²( 1 + 2t) – (1 + 2t) < 0
F(t) = (1 + 2t)(t² – 1) = (1 + 2t)(t – 1)(t + 1) < 0, or back to tan x = t
F(x) = (1 + 2tan x)(tan x – 1)(tan x + 1) = f(x).(g(x).f(x) < 0
(Common period 180⁰)
1. Solve f(x) = 1 + 2tan x = 0. That gives tan x = - 1/2 = tan (- 26⁰56) = tan 153⁰43.
There is discontinuity point at x = 90⁰.
Find the sign status of f(x) by selecting point (x = 45⁰). We get: f(45) = 2(1) + 1 = 3 > 0
Therefor, f(x) > 0 inside the interval (0, 90⁰). Color it red and color the 2 other arc
lengths..
2. Solve g(x) = tan x – 1 = 0. This gives tan x = 1 = tan 45⁰. There is discontinuity point
at x = 90⁰.
Select as checkpoint point (x = 20⁰). We get g(20) = 2(0.36) – 1 < 0. Therefore, g(x) is
negative inside the interval (0, 45⁰). Color it blue.
3. Solve h(x) = tan x + 1 = 0. This gives tan x = -1 --→ x = 135⁰. Discontinuity at 90⁰
Select check point (20⁰). We get h(20) = 2(0.36) + 1 > 0. Therefore, h(x) > 0 inside the
interval (0, 90⁰). Color it red.
By superimposing, the solution set of F(x) < 0 are the 3 open intervals:
(0, 45⁰), and (90⁰, 135⁰), and (153⁰43, 180⁰). See Figure 5.
Fast check by calculator.
F(x) = (1 + 2tan x)(tan² x – 1) < 0
x = 40⁰ --→ F(40) = (1 + 0.84)(0.70 – 1) < 0. Proved.
x = 100⁰ --→ F(100) = (1 – 5.67)(32.16 – 1) < 0. Proved
x = 170⁰ --→ F(170) = (1 – 0.17)(0.03 – 1) < 0. Proved
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Figure 5
Exercise 6. Solve
Figure 6
sin 3x – sin x > cos 3x – cos x
Solution. Use the trig identities (sin a – sin b) and (cos a – cos b) to transform the
inequality:
sin 3x – sin x > cos 3x – cos x
2cos 2x.sin x > - 2sin 2x.sin x
F(x) = 2sin x.(cos 2x + sin 2x) > 0
1. Solve f(x) = sin x = 0. The function sin x is positive (> 0) inside the interval
(0, Ꙥ). Color it red and the other half circle blue.
2. Solve (sin 2x + cos 2x) = 0. Use trig identity: (sin a + cos a) = √2cos (a + Ꙥ/4).
We get: (sin 2x + cos 2x) = √2.cos (2x - Ꙥ/4) = 0. This gives 2 solutions:
a. 2x – Ꙥ/4 = Ꙥ/2 + 2kꙤ --→ 2x = 3Ꙥ/4 + 2kꙤ --→ x = 3Ꙥ/8 + kꙤ
b. 2x – Ꙥ/4 = 3Ꙥ/2 + 2kꙤ --→ 2x = 7Ꙥ/4 + 2kꙤ --→ x = 7Ꙥ/8 + kꙤ
For k = 0, and k = 1, there are 4 end points at: (3Ꙥ/8), (7Ꙥ/8), (11Ꙥ/8), and (15Ꙥ/8).
There are 4 arc lengths. Find the sign status of g(x) by selecting the check point (0).
We get g(0) = (sin 0 + cos 0) = 1 > 0. Therefor, g(x) is positive (> 0) inside the interval
(- Ꙥ/8, 3Ꙥ/8). Color it red and color the 3 other arc lengths.
By superimposing, we see that the solution set of F(x) > 0 are the 3 open intervals;
(0, Ꙥ/8) and (7Ꙥ/8, Ꙥ) and (11Ꙥ/8, 15Ꙥ/8). See Figure 6.
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Check.
F(x) = sin x.cos (2x – Ꙥ/4) > 0
x = Ꙥ/4 --→ F(Ꙥ/4) = sin Ꙥ/4.cos (Ꙥ/2 – Ꙥ/4) = 1/2 > 0. Proved
x = Ꙥ/2 --→ F(Ꙥ/2) = sin Ꙥ/2.cos (3Ꙥ/4) = (+)(-) < 0. Proved
x = 3Ꙥ/2 --→ F(3Ꙥ/2) = (-1)cos (3Ꙥ -Ꙥ/4) = (-1)cos (3Ꙥ/4) = (-)(-) > 0. Proved
Exrcise 7. Solve
tan³ x + tan² x > 3tan x + 3
Solution. Put (tan x + 1) in common factor.
tan² x(tan x + 1) > 3(tan x + 1). We get in standard form:
F(x) = tan² x.(tan x + 1) – 3(tan x + 1) > 0
F(x) = (tan x + 1)(tan² x – 3) = (tan x + 1)(tan x - √3)(tan x + √3) > 0
1. Solve f(x) = tan x + 1 = 0. This gives tan x = -1 --→ x = 3Ꙥ/4
There is discontinuity at x = Ꙥ/2. Find the sign status of f(x) by selecting the check
point x = Ꙥ/4. We get: f(Ꙥ/4) = tan Ꙥ/4 + 1 > 0. Then, f(x) is positive inside the interval
(0, Ꙥ/2). Color it red and color the 2 other arc lengths.
2. Solve g(x) = tan x - √3. This gives: tan x = √3 = tan Ꙥ/3. Discontinuity at x = Ꙥ/2.
Select check point (x = Ꙥ/4). We have: g(Ꙥ/4) = tan Ꙥ/4 - √3 < 0. Then, g(x) < 0 inside
the arc length (0, Ꙥ/2). Color it blue.
3. Solve h(x) = tan x + √3. This gives: tan x = √3 = tan (2Ꙥ/3). Discontinuity at Ꙥ/2.
Select (x = Ꙥ/4) as check point. We get: h(x) = tan Ꙥ/4 + √3 > 0. Then, h(x) > 0 inside
the arc length (0, Ꙥ/2). Color it red
The solution set of F(x) > 0 are the 2 open intervals
(Ꙥ/3, Ꙥ/2) and (2Ꙥ/3, 3Ꙥ/4). See Figure 7
Fast check by calculator.
x = 80⁰ --→ F(80) = (5.67 + 1)(5.67 - √3)(5.67 + √3) > 0 Proved
x = 130⁰ --→ F(130) = (-1.19 + 1)(- 1/19 - √3)(-1.19 + √3) > 0. Proved
Exercise 8. Solve
F(x) = (sin 2x)/(1 - √2cos x) > 0.
Solution. We have the case F(x) = f(x)/g(x).
1. Solve f(x) = sin 2x = 0. This gives 3 solutions:
sin 2x = 0 + 2kꙤ. This gives x = 0 + kꙤ
sin 2x = Ꙥ + 2kꙤ. This gives x = Ꙥ/2 + kꙤ
sin 2x = 2Ꙥ + 2kꙤ. This gives x = Ꙥ + kꙤ
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Figure 7
Figure 8
For k = 0 and k = 1, there are 4 end points at: 0, Ꙥ/2, Ꙥ, and 3Ꙥ/2. There are 4 arc
lengths. f(x) = sin 2x is positive inside the arc length (0, Ꙥ/2). Color it red and color the
3 other arc lengths.
2. Solve g(x) = 1 - √2cos x. This gives cos x = √2/2 = cos (±Ꙥ/4). There are 2 arc
lengths. Select point (0) as check point. We get: g(0) = 1 - √2 < 0. Therefor, g(x) is
negative (< 0) inside the interval (- Ꙥ/4, Ꙥ/4). Color it blue.
The solution set of F(x) > 0 are the 3 open intervals:
(Ꙥ/4, Ꙥ/2), and (Ꙥ, 3Ꙥ/2) and (7Ꙥ/4, 2Ꙥ). See Figure 8.
NOTE. The Nghi Nguyen method also works well when the common period is different
to 2Ꙥ.
Exercise 9. Solve
cos x – sin x.tan x + 1 < 0
Solution. Replace tan x by (sin x/cos x), and 1 by cos x/cos x, then proceed the
transformation:
F(x) = cos² x – sin² x + cos x/cos x < 0
F(x) = (cos 2x + cos x)/cos x < 0
(Trig identity: cos² x – sin² x = cos 2x)
F(x) = [f(x).g(x)]/h(x) = (2cos 3x/2.cos x/2)/cos x < 0
(Trig identity: cos a + cos b)
1. Solve f(x) = cos (3x)/2 = 0. The common period is 4Ꙥ. There are 2 solutions:
(3x)/2 = Ꙥ/2. This gives: 3x = Ꙥ, --→ x = Ꙥ/3 + (4kꙤ)/3
(3x)/2 = 3Ꙥ/2. This gives: 3x = 3Ꙥ --→ x = Ꙥ + (4kꙤ)/3
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For k = 0, k = 1, and k = 2, we get 6 arc lengths and 6 end points at:
(Ꙥ/3), (Ꙥ), (5Ꙥ/3), (7Ꙥ/3), (9Ꙥ/3), (11Ꙥ/3) inside the common period 4Ꙥ.
To find the sign status of f(x) = cos (3x)/2, select the check point (x = 0).
We get f(0) = 1 > 0. So, f(x) is positive (> 0) inside the arc length (11Ꙥ/3, Ꙥ/3) or
(-Ꙥ/3, Ꙥ/3). Color it red and color the other arc lengths.
2. Solve g(x) = cos x/2 = 0. There are 2 solutions:
x/2 = Ꙥ/2 + 2kꙤ. This gives x = Ꙥ + 4kꙤ
x/2 = 3Ꙥ/2 + 2kꙤ. This gives x = 3Ꙥ + 4kꙤ.
There are 2 end points at: (Ꙥ) and (3Ꙥ) and 2 arc lengths for (0, 4Ꙥ).
Select the check point (x = 0). We get g(0) = cos 0 = 1 > 0. Therefore, g(x) is positive
(> 0) inside the half circle (3Ꙥ, 5Ꙥ). Color it red and color the other half circle blue.
3. Solve h(x) = cos x = 0. The function h(x) is positive (> 0) inside the interval
(-Ꙥ/2, Ꙥ/2) and (3Ꙥ/2, 5Ꙥ/2) within the period 4Ꙥ. Color them red and the 2 others
blue.
By superimposing, we see that the combined solution set of (F(x) < 0 are the 4 open
intervals (Ꙥ/3, Ꙥ/2) and (3Ꙥ/2, 5Ꙥ/3) and (7Ꙥ/3, 5Ꙥ/2), and (7Ꙥ/2, 11Ꙥ/3). Figure 9.
Note. There are 4 discontinuities at (Ꙥ/2), (3Ꙥ2), (5Ꙥ/2), and (7Ꙥ/2) when cos x = 0
Figure 9
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Figure 10
Exercise 10. Solve
5sin x + 2cos x > 3
Solution. Divide both sides by 5, we get:
F(x) = sin x + (2/5)cos x – 0.6 > 0
Call t that tan t = sin t/cost = 0.4 --→ t = 21⁰80, and cos t = 0.93
F(x) = sin x + (sint/cost).cos x - 0.6.> 0
F(x) = sin x.cos t + sin t.cos x – 0.56 > 0
F(x) = sin (x + t) – 0.56 > 0
(0.6*cos t= 0.56)
Solve F(x) = sin (x + 21⁰80) – 0.56 = 0. We get:
sin (x + 21⁰80) = 0.56 = sin 33⁰92 = sin (146⁰08). This leads to:
x + 21⁰80 = 33⁰92 --→ x = 33.92 – 21.80 = 12⁰12
x + 21.80 = 146.08 --→ x = 146.08 – 21.80 = 124⁰28.
There are 2 end points at (12⁰12) and (124⁰08), and 2 arc lengths. To find the sign
status of F(x), select the check point (x = 90⁰). We get: F(90⁰) = 5 + 0 – 3 = 2 > 0.
Therefore, F(x) is positive (> 0) inside the interval (12⁰12, 124⁰28). That is the solution
set of the trig inequality. See Figure 10.
Exercise 11. Solve
tan x + sec x > 1
(Common period 2Ꙥ)
Solution. Replace 1 by (cos x/cos x) and move it to the left side, we get:
F(x) = sin x/cos x + 1/cos x – cos x/cos x > 0
F(x) = (sin x + 1 – cos x)/cos x = f(x)/g(x) > 0
1. Solve f(x) = sin x – cos x + 1 = 0. Using trig identity: sin x – cos x = √2sin (x + Ꙥ/4),
we get:
f(x) = √2sin (x + Ꙥ/4) + 1 = 0.
This gives: sin (x + Ꙥ/4) = - 1/√2 = √2/2 = sin (5Ꙥ/4) = sin (7Ꙥ/4)
a. x + Ꙥ/4 = 5Ꙥ/4 -→ x = 5Ꙥ/4 – Ꙥ/4 = Ꙥ
b. x + Ꙥ/4 = 7Ꙥ/4 --→ x = 7Ꙥ/4 – Ꙥ/4 = 3Ꙥ/2
There are 2 end points at (Ꙥ) and (3Ꙥ/2) and 2 arc lengths. Find the sign status of f(x)
by selecting the point (x = 5Ꙥ/4). We get: f(5Ꙥ/4) = -√2/2 - √2/2 + 1 < 0. Therefor, f(x) is
negative (< 0) inside this arc length. Color it blue, and color the rest red.
Note. F(x) is undefined at (x = Ꙥ/2) and (x = 3Ꙥ/2) where cos x = 0
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2. g(x) = cos x = 0. By definition, g(x) = cos x is positive inside the interval (-Ꙥ/2, Ꙥ/2)
Color it red and the rest blue.
By superimposing, we see that the solution set of F(x) > 0 are the 2 open intervals:
(Ꙥ, 3Ꙥ/2), and (-Ꙥ/2, Ꙥ/2). See Figure 11.
Check
x = Ꙥ/4 --→ F(Ꙥ/4) = (√2/2 + √2/2 + 1)/√2/2 > 0. Proved
x = 3Ꙥ/4 --→ F(3Ꙥ/4) = (√2/2 - √2/2 + 1)/-√2/2 = (+/-) < 0. Proved.
Exercise 12. Solve
2sin² x- 4cos² x + 2sin 2x > 1
Solution. Divide both side by cos² x, and call t = tan x, we get
2t² – 4 + 4t > (1/cos² x) = t² + 1
(trig identity: tan² x + 1 = 1/cos² x)
F(x) = t² + 4t – 5 > 0
F(x) = tan² x + 4tan x – 5 > 0
(Common period 180⁰)
Solve F(x) = 0. Since a + b + c = 0, this quadratic equation in tan x has 2 real roots:
tan x = 1 and tan x = - 5. F(x) can be factored into:
F(x) = (tan x – 1)(tan x + 5) > 0
1. Solve f(x) = tan x – 1 = 0. This gives tan x = 1 --→ x = 45⁰.
f(x) is negative (< 0) inside the interval (0, 45⁰). Color it blue and color the 2 other arc
lengths. There is discontinuation at x = 90⁰
2. Solve g(x) = tan x + 5 = 0. This gives: tan x = - 5 --→ x = - 78⁰69, or x = 101⁰31.
The function g(x) > 0 inside the arc length (0⁰, 90⁰). Color it red and color the 2 other
arc lengths.
By superimposing we see that the combines solution set of F(x) > 0 are the 2 open
intervals:
(45⁰, 90⁰) and (90⁰, 101⁰31). See Figure 12.
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Figure 11
Figure 12
ADVANTAGES OF THE NGHI NGUYEN METHOD
1. The 2 extremities (0 and 2Ꙥ) of the sign chart are separate. In the new method, the
two extremities joint together and show the periodic property of trig functions.
2. On the sign chart, all x-values from all operations are only placed on the first line. In
case of complex trig inequalities such as F(x) = f(x).g(x).h(x) ≥ 0 (or≤ 0), the high
number of x makes the chart stuffy and confusing. In the new method, the number of x
only depends on, in each step, the solving of only one trig inequality.
3. The simple rules of end points and arc lengths make the new method more
convenient than the sign chart method.
4. Applying the number line checking method on the number unit circle is a fine
technique, that is as good as using the triple unit circle to solve complex trig
inequalities.
(This article was authored by Nghi H Nguyen, Updated Mar. 01, 2021)
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