SOLVING COMPLEX TRIG INEQUALITIES - Showcase Examples.
(Authored by Nghi H Nguyen – March 06, 2021)
This article shows a few selected examples of solving complex trig inequalities, by
using the Nghi Nguyen method (Google, Bing, Yahoo Search)
Exercise 1. Solve
sin 2x/(3 – 2tan x) < 0
Solution. We have the inequality in the form: F(x) = f(x)/g(x) < 0
(Common period Ꙥ)
1. Solve f(x) = sin 2x = 0. This gives 2 solutions:
a. sin 2x = 0 + 2kꙤ --→ x = 0 + kꙤ
b. sin 2x = Ꙥ + 2kꙤ --→ x = Ꙥ/2 + kꙤ
c. 2x = 2Ꙥ + 2kꙤ ---→ x = Ꙥ + 2kꙤ
There are 3 end points at: 0, Ꙥ/2, Ꙥ, and 2 equal arc lengths for (0, Ꙥ)
Find the sign status (+ or -) of f(x), by selecting the point (x = Ꙥ/4). We have
f(Ꙥ/4) = sin Ꙥ/2 = 1 > 0. Therefore, f(x) > 0 inside the arc length (0, Ꙥ/2). Color it red
and color the other blue.
2. Solve g(x) = (3 – 2tan x) = 0. This gives: tan x = 3/2 = tan 56⁰31. There is one
discontinuity at (x = Ꙥ/2) and one end point at (x = 56⁰31) and 3 arc lengths. Select
point (45⁰) as check point. We get: f(45) = 3 – 2*1) = 1 > 0. Therefor, f(x) is positive
inside arc length (0, 56.31). Color it red and color the 2 other arc lengths. Figure 1
The solution set of F(x) < 0 are the 2 open interval (56⁰31, 90⁰) and (90⁰, 180⁰)
Figure 1
Figure 2
Check by calculator
x =45⁰. This gives F(1/(3 – 2) = 1/1 > 0. Proved
x = 80⁰. This gives: F(80) = sin 160/(3 – 2*5.67) = 0.34/-8.34 < 0. Proved
x = 150⁰. This gives: F(150) = sin 300/(2 – 3.tan150) = - 0.87/(2 – 3*0.57) < 0. Proved
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Exercise 2. Solve
sin 2x + √3.cos 2x > √2
(period Ꙥ)
Solution. Call t the arc that tan t = √3. This gives: t = Ꙥ/3, and cos t = √2/2.
Solve F(x) = sin 2x + (sin t/cos t).cos 2x - √2 = 0
Applying the trig identity (sin (a + b)), we get:
F(x) = sin (2x + t) - √2.cos t = 0. This gives:
sin (2x + Ꙥ/3) = √2(1/2) = √2/2 = sin Ꙥ/4. This leads to:
2x + Ꙥ/3 = Ꙥ/4 + 2kꙤ. This gives: 2x = - Ꙥ/12 + 2kꙤ --→ x = - Ꙥ/24 + kꙤ
2x + Ꙥ/4 = 3Ꙥ/4 + 2kꙤ. This gives: 2x = Ꙥ/2 + 2kꙤ --→ x = Ꙥ/4 + kꙤ
There are two end points at (Ꙥ/4) and (-Ꙥ/24) or (23Ꙥ/24) for (0, Ꙥ). There are 3 arc
lengths. Select point (x = Ꙥ/2) as check point.
We get: F(Ꙥ/2) = sin Ꙥ + √3.cos Ꙥ - √2 = - √3 -√2 < 0. There for F(x) is negative (< 0)
inside the arc length (Ꙥ/4, 23Ꙥ/24). See Figure 2.
The solution set of F(x) > 0 are the 2 open intervals (0, Ꙥ/4) and (23Ꙥ/24, Ꙥ)
Check by calculator.
sin 2x + √3.cos 2x - √2 > 0
x = 40⁰ --→ F(40) = sin 80 + √3cos 80 - √2 = 0.98 + 1.69 – 1.41 > 0. Proved.
x = 90⁰ --→ F(90) = sin 180 + √3cos 180 - √2 = - √3 - √2 < 0. Proved
x = 175⁰ --→ F(175) = sin 350 + √3cos350 - √2 = - 0.17 + 1.70 – 1.41 > 0. Proved
Exercise 3. 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 + 4kꙤ --→ x = Ꙥ/3 + (4kꙤ)/3
(3x)/2 = 3Ꙥ/2. This gives: 3x = 3Ꙥ + 4kꙤ --→ x = Ꙥ + (4kꙤ)/3
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.
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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). See Figure 3.
Note. There are 4 discontinuities at (Ꙥ/2), (3Ꙥ2), (5Ꙥ/2), and (7Ꙥ/2) when cos x = 0
Figure 3
Figure 4
Check.
F(x) = (cos 2x + cos x)/cos x< 0
x = Ꙥ/4 --→ F(Ꙥ/4) = 0 + 1/1 > 0. Proved
x = 3Ꙥ/4 --→ F(3Ꙥ/4) = (0 + √2/2)√2/2 > 0. Proved
x = 280⁰ --→ F(280) = (cos 560 + cos 280)cos 280 = (-0.94 + 017)/0.17 < 0. Proved
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Exercise 4. 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 4.
Exercise 5. 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
(sin x – cos x + 1)/cos x
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.
x = 5Ꙥ/4 --→ F(5Ꙥ/4) = (√2/2 + √2/2)√2/2 > 0. Proved
Exercise 6. 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 > 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 6.
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Figure 5
Exercise 7. Solve
Figure 6
3(sin x + cos x) + 2sin 2x + 3 < 0 (1)
(period 2Ꙥ)
Solution. Call (sin x + cos x) = t, we get: t² = 1 + 2sin x.cos x = 1 + sin 2x.
sin 2x = t² - 1. The inequality (1) becomes: F(x) = 2t² + 3t + 1 < 0.
Solve F(x) = 2t² + 3t + 1 = 0
This is a quadratic equation in t. Since a - b + c = 0, the 2 real roots are: t = - 1 and
t = -c/a = - 1/2. Put back t = sin x + cos x = √2.sin (x + Ꙥ/4) in the inequality, we get the
factoring form:
F(x) = f(x).g(x) = [√2.sin (x + Ꙥ/4) + 1][√2.sin (x + Ꙥ/4) + 0.5 ] < 0
1. Solve f(x) = √2.sin (x + Ꙥ/4) + 1 = 0. This gives: sin (x + Ꙥ/4) = - √2/2 .
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). There are 2 arc lengths. Select point (5Ꙥ/4)
as check point. We get f(5Ꙥ/4) = √2.sin (3Ꙥ/2) + 1 = -√2 + 1 < 0. Therefore, f(x) < 0
inside the interval (Ꙥ, 3Ꙥ/2). Color it blue, and color the rest red.
2. Solve g(x) = √2.sin (x + Ꙥ/4) + 0.5 = 0. This gives: sin (x + Ꙥ/4) = - √2/4.
x + 45⁰ = 200⁰70 --→ x = 200.70 – 45 = 155⁰70
x + 45 = 329⁰30 --→ x = 329.30 – 45 = 294⁰30
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There are 2 end points at (155⁰70) and (294⁰30), and 2 arc lengths. Select point (180⁰)
as check point. We get: g(180) = √2.sin (225) + 0.5 = - 1 + 0.5 < 0. Therefore, g(x) is
negative inside the arc length (155⁰70, 294⁰30).
We may change the answer to radians by using the rule of conversion.
x = 155⁰70. Convert to radian: x (rad.) = 155.70 x Ꙥ/180 = 0.87Ꙥ
x = 294⁰30 . Convert to radians: x (rad.) = 294.30 x Ꙥ/180 = 1. 64Ꙥ
The solution set of F(x) < 0 are the 2 open intervals
(0.87Ꙥ, Ꙥ) and (1.5Ꙥ, 1.64Ꙥ). See Figure 7.
Figure 7
Exercise 8. Solve
Figure 8
1 + tan 2x = (1 – sin 2x)/cos² 2x
Solution. F(x) = cos² 2x + (sin 2x/cos 2x)(cos² 2x) + sin 2x - 1 > 0
F(x) = cos² 2x + sin 2x.cos 2x + sin 2x - 1 > 0
F(x) = sin 2x(cos 2x + 1) + cos² 2x – 1 = > 0
(put sin 2x in common factor)
F(x) = sin 2x(cos 2x + 1) + (cos 2x – 1)(cos 2x + 1) > 0
F(x) = f(x).g(x) = (cos 2x + 1)(sin 2x + cos 2x - 1) > 0 (put (2x + 1) in common factor)
1. Solve f(x) = cos 2x + 1 = 0. This function is always positive regardless of x (except
when x = Ꙥ/2). The sign status of F(x) is the sign status of g(x).
2. Solve g(x) = sin 2x + cos 2x – 1 = 0.
Use trig identity: sin a + cos a = √2.sin (a + Ꙥ/4), we get:
g(x) = √2.sin (2x + Ꙥ/4) – 1 = 0 --->: sin (2x + Ꙥ/4) = √2/2 = sin (Ꙥ/4) = sin (3Ꙥ/4).
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a. 2x + Ꙥ/4 = Ꙥ/4 --→ x = 0 + kꙤ
b. 2x + Ꙥ/4 = 3Ꙥ/4 --→ x = Ꙥ/4 + kꙤ
For k = 0 and k – 1, there are 4 end points at (0), (Ꙥ/4), (Ꙥ), (5Ꙥ/4), and 4 arc lengths.
Select point (Ꙥ/2) as check point. We get: g(Ꙥ/2) = sin Ꙥ + cos Ꙥ – 1 = 0 - 1 - 1 < 0.
Therefor, g(x) is negative (< 0) inside the interval (Ꙥ/4, Ꙥ).
Color it blue and color the other 3 arc lengths
The solution set of F(x) > 0 are the 2 open intervals: (0, Ꙥ/4) and (Ꙥ, 5Ꙥ/4). Figure 8.
Check.
F(x) = (cos 2x + 1)(sin 2x + cos 2x - 1) > 0
x = 30⁰ → F(30) = (cos 60 + 1)(sin 60 + cos 60 – 1) = 1.5 (0.87 + 0.5 – 1) > 0. Proved.
X = 80⁰ --→ F(80) = (cos 160 + 1)(sin 160 + cos 610 – 1) = 0.06(0.34 - 0.94 – 1) < 0.
Proved
Exercise 9. Solve
1/(cos x) < 4sin x + 6cos x
(period 2Ꙥ)
Solution. Divide both side by cos x (condition cos x ≠ 0; x ≠ Ꙥ/2; x ≠ 3Ꙥ/2).
1/(cos² x) = 4 tan x + 6
1 + tan² x = 4tan x + 6
(use trig identity: 1 + tan² x = 1/cos² x)
F(x) = tan² x – 4tan x – 5 < 0.
This is a quadratic equation in tan x. Since a – b + c = 0, the 2 real roots are: tan x = 1, and tan x = -c/a = 5. Write F(x) in factoring form:
F(x) = f(x).g(x) = (tan x + 1)(tan x – 5) < 0
1. Solve f(x) = tan x – 1. This gives: tan x = -1 --→ x = -Ꙥ/4 + kpi.
For k = 0 and k = 1, there are 2 end points at (-Ꙥ/4), and (3Ꙥ/4), and 4 arc lengths for
(0, 2Ꙥ). There are 2 discontinuities at (x = Ꙥ/2) and (x = 3Ꙥ/2).
To find the sign status of f(x) select point (x = Ꙥ). We get f(Ꙥ) = tan Ꙥ + 1 = 0 + 1 > 0.
Therefore, f(x) is positive (> 0) inside the arc length (Ꙥ/2, 5Ꙥ/4). Color it red and color
the 3 other arc lengths.
2. Solve g(x) = tan x – 5 = 0.. This gives: tan x = 5 --→ x = 78⁰69 or x = 0.44Ꙥ + kꙤ.
There are 2 end points at (0.44Ꙥ) and (1.44Ꙥ). There are 4 arc lengths.
Select point (x = Ꙥ) as check point. We get: g(Ꙥ) = 0 – 5 < 0. Therefor, g(x) is
negative inside the arc length (0.44Ꙥ, 1.44Ꙥ). Color it blue and color the 3 other arc
lengths.
By superimposing, the solution set of F(x) > 0 are the 2 open intervals:
(0.44Ꙥ, 3Ꙥ/4) and (1.44Ꙥ, 7Ꙥ/4). See Figure 9.
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Check
F(x) = tan² x – 4tan x – 5 < 0
x = 0 --→ F(0) = 0 – 0 – 5 < 0. Proved
x = Ꙥ/4 --→ F(Ꙥ/4) = 1 – 4 – 5 < 0. Proved
x = 5Ꙥ/4 --→ F(5Ꙥ/4) = 1 – 4 – 5 < 0. Proved
Figure 9
Exercise 10. Solve
Figure 10
cos² 2x + 7sin² x > 2
(Common period 2Ꙥ)
Solution: Using trig identity: (1 – cos 2a = 2sin²a), we get:
F(x) = cos² 2x + (7/2)(1 – cos 2x) – 2 > 0
F(x) = 2cos² 2x – 7cos 2x + 3 > 0
This is a quadratic equation in cos 2x. After factoring, we get:
F(x) = f(x).g(x) = (2cos 2x – 1)(cos 2x - 3) > 0
1. Solve f(x) = 2cos 2x – 1 = 0. This gives: cos 2x =1/2 = cos ± Ꙥ/3 + 2kꙤ --->
x = ± Ꙥ/6 + kꙤ. There are 4 end points at (± Ꙥ/6) and (± 7Ꙥ/6). There are 4 arc
lengths. Select check point (x = 0). We get f(0) = 2 – 1 = 1 > 0. Therefore, f(x) is
positive (> 0) inside the arc length (- Ꙥ/6, Ꙥ/6). Color it red and color the 3 other arc
lengths. See Figure 10.
2. g(x) = cos 2x – 3 is always negative regardless of x. So, the sign status of F(x) is
the opposite one of f(x).
In the Figure 10, the solution set of F(x) > 0 is the opposite of the solution set of
f(x) < 0. They are the 2 blue open intervals: (Ꙥ/6, 5Ꙥ/6) and (7Ꙥ/6, 11Ꙥ/6).
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Check.
cos² 2x + 7sin² x – 2 > 0
x = Ꙥ/2 --→ F(Ꙥ/2) = 1 + 7 = 8 – 2 > 0. Proved
x = 3Ꙥ/2 --→ F(3Ꙥ/2) = 1 + 7 = 8 – 2 > 0. Proved
x = 0 --→ F(0) = 1 + 0 – 2 < 0. Proved.
x = Ꙥ --→ F(Ꙥ) = 1 – 2 < 0. proved
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 basic 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. 06, 2021)
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