SOLVING COMPLEX TRIG INEQUALITIES - EXERCISES
(by Nghi H Nguyen– Jan. 15, 2021)
GENERALITIES ON THE METHOD.
The author of this article uses Nghi H Nguyen’s method to solve some selected trig exercises.
This innovative and visual method, recently posted on Google, solves complex trig inequalities
by using the number unit circle that can be numbered in radians or degrees.
A complex trig inequality F(x) can be transformed into many basic 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)
First, this method finds the sign status (+ or -) of one trig basic function f(x) when the variable
arc x varies inside various arc lengths on the unit circle. Then, it reports the sign status (+ or -)
of f(x) on one unit circle that is numbered (in radians or degrees) following the common period.
Next, it colors the arc lengths, red for f(x) positive, and blue when f(x) < 0.
Do the same thing for g(x), and report its sign status on a second concentric unit circle.
By superimposing, we easily see the combined solution set for F(x) = f(x).g(x) ≤ 0 (or ≥ 0).
Students can also use the triple unit circle to solve complex trig inequalities in the form
F(x) = f(x).g(x).h(x).
In order to solve complex trig inequalities, students must master these domains:
- Solving basic trig equations (or similar) by using the unit circle and the 4 axis (cos, sin, tan
and cot).
Examples. Solve sin x > 0.5
cos (x – Ꙥ/3) ≥ √2/2
tan (x – 30⁰) > 1
cot (x – Ꙥ/4) > 0.5 cos (2x/3 – Ꙥ/3) < 0.75
sin (x/2 + Ꙥ/4) < 1
- Transformation of a complex trig inequality F(x) into one or many basic trig inequalities in
the form F(x) = f(x).g(x) < 0 (or > 0) or F(x) = f(x).g(x).h(x) < 0 (or > 0). Transformation
means include using trig identities and common algebraic manipulations. There are about 35
trig identities that are usually shown in trig books.
- A few simple rules for end points and arc lengths:
- For f(x) = sin x and f(x) = cos x, and similar, there are 2 endpoints and 2 arc lengths for (0, 2Ꙥ)
- For f(x) = sin 2x & f(x) = cos 2x, and similar, there are 4 endpoints and 4 arc lengths for (0, 2Ꙥ)
- For f(x) = sin 3x & f(x) = cos 3x, and similar, there are 6 endpoints and 6 arc lengths (0, 2Ꙥ)
- For f(x ) = tan x there are 1 endpoint, one discontinuation and 3 arc lengths for (0, Ꙥ)
- For f(x) = cot x, there are 1 endpoint, one discontinuation, and 3 arc lengths for (Ꙥ/2, 3Ꙥ/2)
Page 1 of 9
SOLVING BASIC TRIG INEQUALITIES, OR SIMILAR.
Exercises 1. Solve cos (2x – Ꙥ/6) > 0
Solution.
F(x) = cos (2x – Ꙥ/6) > 0
First, solve F(x) = cos (2x – Ꙥ/6) = 0.
(0, 2Ꙥ)
There are 2 solutions:
a. (2x – Ꙥ/6) = Ꙥ/2 + 2kꙤ --> 2x = Ꙥ/2 + Ꙥ/6 + 2kꙤ = 4Ꙥ/6 = 2Ꙥ/3 + 2kꙤ
That gives: x = Ꙥ/3 + kꙤ
b. (2x – Ꙥ/6) = 3Ꙥ/2 + 2kꙤ → 2x = 3Ꙥ/2 + Ꙥ/6 + 2kꙤ = 5Ꙥ/3 + 2kꙤ
x = (5Ꙥ/6) = kꙤ
There are 4 arc lengths and 4 endpoints at: Ꙥ/3; 5Ꙥ/6; 4Ꙥ/3; 11Ꙥ/6
To find the sign status, select x = 0 as the check point. We get F(0)= cos (0 – Ꙥ/6) =
= cos (-Ꙥ/6) > 0. Color this arc length red, and color the 3 other arc lengths following the
property of the end point.
We see that the solution set F(x) > 0 are the 2 open intervals (red):
(11Ꙥ/6, Ꙥ/3) and (5Ꙥ/6, 4Ꙥ/3)
Figure 1
Exercises 2. Solve
sin (2x + Ꙥ/3) < 0
Solution.
F(x) = sin (2x + Ꙥ/3) < 0
sin (2x + Ꙥ/3) = 0 when:
2x + Ꙥ/3= 0 → 2x = - Ꙥ/3 + 2kꙤ → x = - Ꙥ/6 + kꙤ, or x = 11Ꙥ/6 + kꙤ
Page 2 of 9.
2x + Ꙥ/3 = Ꙥ → 2x = Ꙥ – Ꙥ/3 + 2kꙤ = 2Ꙥ/3 + 2kꙤ → x = Ꙥ/3 + kꙤ
2x + Ꙥ/3 = 2Ꙥ → 2x = 2Ꙥ – Ꙥ/3 + 2kꙤ = 5Ꙥ/3 + 2kꙤ → x = 5Ꙥ/6 + kꙤ
For k = 0 and k = 1 There are 4 end points at: x = Ꙥ/3, 5Ꙥ/6 , 4Ꙥ/3, and 11Ꙥ/6
There are 4 arc lengths. Select x = 0 as check point. We get F(0) = sin (0 + Ꙥ/3) > 0 (red)
Color the 3 other arc lengths.
Finally, the solution set of F(x) < 0 are the 2 open intervals (blue):
(Ꙥ/3, 5Ꙥ/6) and (4Ꙥ/3, 11Ꙥ/6). See Figure 2
Figure 2
Exercise 3. Solve
Solution:
tan (x – 30⁰) > 1
F(x) = tan (x – 30⁰) - 1 > 0
(Period 180⁰)
F(x) = tan (x – 30⁰) -1 > 0 when the arc (x – 30⁰) varies inside the arc length (45⁰, 90⁰)
45⁰ < x – 30⁰ < 90⁰
75⁰ < x < 120⁰
There is a discontinuity at x = 90⁰. Plot these 2 end points on the circle. They divide
the circle into 3 arc lengths.
Inside the arc length (0, 75⁰), select the check point x = 50⁰, we get F(x) = tan (20⁰) – 1 < 0,
then F(x) < 0 . Color it blue.
Inside the arc length (75⁰, 120⁰), F(x) > 0 following the property of end point. Color it red.
Inside the arc length (120⁰, 180⁰), F(x) < 0. Color it blue.
Page 3 of 9
The solution set of F(x) > 0 is the open interval (75⁰, 120⁰).
Figure 3
Exercise 4. Solve
F(x) = (1 + sin x)/cos x ≤ 1.
Solution. F(x) = (1 + sin x)/cos x – cos x/cos x = (1 + sin x - cos x)/cos x < 0
We have the form F(x) = f(x)/g(x) < 0
(Common period 2Ꙥ)
1. First, solve f(x) = 1 + sin x – cos x = 0
Use the trig identity (sin x – cos x) = - √2. cos (x + Ꙥ/4), we get:
f(x) = 1 - √2.cos (x + Ꙥ/4) ≤ 1. That gives: cos (x + Ꙥ/4) = √2/2 = cos Ꙥ/4 = cos 7Ꙥ/4.
a. x + Ꙥ/4 = Ꙥ/4. This gives the end point at: x = 0
b. x + Ꙥ/4 = 7Ꙥ/4 . This gives the end point at: x = 3Ꙥ/2
On the first unit circle, we have 2 end points at (0) and (3Ꙥ/2) and 2 arc lengths (0, 3Ꙥ/2) and
(3Ꙥ/2, 2Ꙥ). There are 2 discontinuities at (Ꙥ/2) and (3Ꙥ/2) when cos x = 0.
Select point (7Ꙥ/4) as check point. We get f(7Ꙥ/4) = 1 - √2 < 0.
Therefor, f(x) is negative (< 0) inside arc length (3Ꙥ/2, 2Ꙥ). Color it blue, and the rest red.
Figure 4
Page 4
2. Next, solve g(x) = cos x = 0.
By definition, we have g(x) > 0, inside arc length (-Ꙥ/2, Ꙥ/2). Color it red and color the rest
blue. Report the sign status of f(x) and g(x) on 2 concentric number unit circles.
By superimposing, we see that the combined solution set is the closed interval (Ꙥ/2, 2Ꙥ/).
Exercise 5. Solve
F(x) = sin 3x – sin x < cos 2x
Solution.
F(x) = sin 3x – sin x – cos 2x < 0
(Common period 2Ꙥ)
Use trig identity (sin a – sin b) (# 29) to transform the equality.
F(x) = 2cos 2x.sin x – cos 2x = cos 2x(2sin x – 1) = f(x).(g(x) < 0
1. Solve f(x) = cos 2x = 0. we have cos 2x = 0 when
a. 2x = Ꙥ/2 + 2kꙤ → that gives x = Ꙥ/4 + kꙤ
b. 2x = 3Ꙥ/2 + 2kꙤ → that gives 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. Find the sign status of each arc length.
Inside arc length (Ꙥ/4, 3Ꙥ/4) , select point (Ꙥ/2) as check point. We have:
f(Ꙥ/2) = cos (Ꙥ) = - 1 < 0. Therefor, f(x) is negative in this arc length. Color it blue.
Color the other intervals.
2. Solve g(x) = 2sin x – 1 = 0. We have sin x = 1/2 that gives 2 end points at :
x = Ꙥ/6, and x = (5Ꙥ/6), and 2 arc lengths. Select the point (Ꙥ/2) as check point
g(Ꙥ/2) = 2sin Ꙥ/2 – 1 = 2 – 1 > 0. Therefor, g(x) is positive inside the interval (Ꙥ/6, 5Ꙥ/6).
Color it red and the rest blue.
Figure 5
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By superimposing we see that the combined solution set are the 3 open intervals:
(Ꙥ/4, 3Ꙥ/4) and (5Ꙥ/6, 5Ꙥ/4) and (7Ꙥ/4, 11Ꙥ/6)
Fast check with calculator.
sin 3x – sin x – cos 2x < 0
x = 20⁰ → F(x) = sin 60⁰ – sin 20⁰ - cos 40⁰ = 0.87 – 0.34 - 0.77 < 0 (proved)
x = 100⁰ → F(x) = sin 300⁰ – sin 100⁰ – cos 200⁰ = - 0.87 - 0.98 + 0.99 < 0 (proved)
x = 200⁰ → F(x) = sin 600⁰ – sin 200⁰ – cos 400⁰ = - 0.87 + 0.34 – 0.76 < 0 (proved)
x = 300⁰ → F(x) = sin 900⁰ – sin 300⁰ – cos 600⁰ = 0 + 0.87 + 0.5 > 0 (proved)
Exercise 6. Solve
sin 3x – sin x < sin 2x
Solution. Transform the inequality by using these trig identities:
sin 2x = 2sin x.cos x, and (sin a – sin b) (# 29), and (cos a – cos b) (# 27).
F(x) = sin 3x – sin x – sin 2x < 0
F(x) = 2cos 2x.sin x – 2sin x.cos x < 0
F(x) = 2sin x(cos 2x – cos x) < 0
F(x) = 2sin x[(- 2sin (3x/2).sin (x/2)] < 0
F(x) = 4.sin (3x/2).sin x.sin (x/2). > 0
F(x) = f(x).g(x).h(x) > 0
(side transposing)
(Common period 4Ꙥ)
1. Solve f(x) = sin (3x/2) within the common period 4Ꙥ. We have sin (3x/2) = 0 when:
3x/2 = 0
3x/2 = Ꙥ
3x/2 = 2Ꙥ
This gives: 3x/2 = 0 + 2kꙤ and x = 0 + (4kꙤ)/3
This gives: 3x/2 = Ꙥ + 2kꙤ and x = 2Ꙥ/3 + (4kꙤ/3)
This gives: 3x/2 = 2Ꙥ + 2kꙤ and x = 4Ꙥ/3 + (4kꙤ/3)
For k = 0, k = 1, and k = 2, we get 6 end points at: 0, 2Ꙥ/3, 4Ꙥ/3, Ꙥ, 8Ꙥ/3 and 10Ꙥ/3.
There are 6 equal arc lengths.
Find the sign status (+ or -) of f(x) inside these 6 arc lengths.
Use (Ꙥ/6) as check point. We get: f(Ꙥ/6) = sin (Ꙥ/4) > 0. Therefor, f(x) = sin (3x/2) > 0 inside
this arc length. Color it red. Then, color the 5 other arc lengths following the property of end
points. See Figure 6.
2. Solve g(x) = sin x = 0.
There are 4 end points at: (0), (Ꙥ), (2Ꙥ), and (3Ꙥ). There are 4 equal arc lengths.
- Inside the arc length (0, Ꙥ), we have g(x) = sin x > 0. Color it red
- Inside the arc length (Ꙥ, 2Ꙥ), we have g(x) < 0. (Blue)
- Inside arc length (2Ꙥ, 3Ꙥ), we have g(x) > 0 (red)
- Inside arc length (3Ꙥ, 4Ꙥ), we have g(x) < 0 (Blue)
Page 6 of 9
3. Solve h(x) = sin x/2 = 0
There are 2 end points at: (0), and (2Ꙥ). There are 2 equal arc lengths.
Inside arc length (0, 2Ꙥ), we have h(x) > 0. Color it red
Inside arc length (2Ꙥ, 4Ꙥ), we have h(x) < 0. Color it blue.
Figure 6
By superimposing, we see that the combined solution set are the 4 open intervals:
(0, 2Ꙥ/3) and (Ꙥ, 4Ꙥ/3) and (2Ꙥ, 8Ꙥ/3) and (3Ꙥ, 10Ꙥ/3)
Check answers.
sin 3x – sin x < sin 2x
x = 7Ꙥ/6 --> We get:
or
x = 9Ꙥ/4 → We get
or
x = 170⁰ → We get
or
sin (3Ꙥ/2) – sin (7Ꙥ/6) < sin (7Ꙥ/3)
- 1 – 0.5 < 0.87 (proved)
sin (3Ꙥ/4) – sin (Ꙥ/4) < sin (Ꙥ/2)
√2/2 - √2 < 1 (proved)
sin 510⁰ – sin 170⁰ < sin 340⁰
0.5 - 0.17 < - 0.34 (not true)
Exercise 7. Solve
tan 2x + tan x < 0
Solution.
F(x) = tan 2x + tan x < 0
Using trig identity (# 30), (tan a + tan b) = sin (a + b)/(cos a.cos b), we get:
F(x) = sin (3x)/(cos x.cos 2x) = f(x)/(g(x).h(x))
Page 7 of 9
(common period 2Ꙥ)
1. Solve f(x) = sin 3x = 0. On the unit circle, we have sin 3x = 0 when:
3x = 0 + 2kꙤ. This gives: x = 0 + (2kꙤ)/3
3x = Ꙥ + 2kꙤ. This gives x = Ꙥ/3 + (2kꙤ)/3
3c = 2Ꙥ + 2kꙤ. This gives x = (2Ꙥ/3) + (2kꙤ)/3
For k = 1, k = 2, and k = 3, there are 6 end points at: (0), (Ꙥ/3), (2Ꙥ/3), (Ꙥ), (4Ꙥ/3), and (5Ꙥ/3).
There are totally 6 arc lengths. Find the sign status of f(x) = sin 3x when x varies inside these 6
arc lengths.
Select (Ꙥ/2) as check point. We get f(x) = sin 3x = sin (3Ꙥ/2) = - 1 < 0. Therefor, f(x) is
negative (-) inside this arc length (Ꙥ/3, 2Ꙥ/3). Color it blue. Color the 5 other arc lengths
following the rule of end point. See Figure 7
2. Solve g(x) = cos 2x. We have cos 2x = 0 when
2x = Ꙥ/2 + 2kꙤ. This gives x = Ꙥ/4 + kꙤ
2x = 3Ꙥ/2 + 2kꙤ. This gives: 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 arc lengths. Find the sign status of g(x) = cos 2x inside arc length (Ꙥ/4, 3Ꙥ/4).
Select point (Ꙥ/2) as check point, we get: g(Ꙥ/2) = cos (Ꙥ) = -1 < 0. Therefor, g(x) < 0 in this
arc length. Color it blue, and color the other arc lengths.
3. Solve h(x) = cos = 0. There are 2 end points at: (Ꙥ/2) and (3Ꙥ/2).
The property of the trig function h(x) = cos x gives the sign status (+) when x varies inside the
arc length (-Ꙥ/2, Ꙥ/2). Color it red and color the half circle blue.
Next, report the sign status of f(x), g(x) and h(x) on 3 concentric unit circles. Figure 7
Page 8 of 9
By superimposing, we see that the solution set of F(x) are the 6 open intervals:
(Ꙥ/4, Ꙥ/3), and (Ꙥ/2, 2Ꙥ/3), and (3Ꙥ/4, Ꙥ), and (5Ꙥ/4, 4Ꙥ/3), and (3Ꙥ/2, 5Ꙥ/3), and
(7Ꙥ/4, 2Ꙥ).
Fast check with calculator.
F(x) = tan 2x + tan x < 0
x = 30⁰ → F(30) = tan 60 + tan 30 = 1.73 + 0.58 > 0 (proved)
x = 50⁰ → F950) = tan 100 + tan 50 = - 5.67 + 1.19 < 0 (proved)
x = 100⁰ → f(100) = tan 200 + tan 100 = 0.36 – 5.67 < 0 (proved)
x = 280⁰ → f(280) = tan 560 + tan 280 = 0.36 – 5.67 < 0 (proved)
x = 340⁰ → F(340) = tan 680 + tan 340 = - 0.84 – 0.37 < 0 (proved)
Page 9 of 9