SOLVING COMPLEX TRIG INEQUALITIES – Method and Examples (Part 2)
(Authored by Nghi H Nguyen, Dec. 01, 2021)
GENERALITIES.
To solve a complex trig inequality F(x) ≤ 0 (or ≥ 0), we transform it into one inequality
containing a few basic 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), or
F(x) = f(x)/g(x) ≤ 0 (or ≥ 0), or
F(x) = (f(x).g(x))/h(x) ≤ 0 (or ≥ 0).
Next, we solve these basic trig inequalities by using concentric numbered unit circles.
Then, by superimposing, the combined solution set of F(x) can be easily seen.
Example 1. Solve
csc 3x > 2
Solution.
1/sin 3x > 2 (1)
General form:
1/sin 3x – 2 > 0 (2)
Replace 2 by 2(sin 3x)/sin 3x. The inequality (2) becomes, with condition sin 3x ≠ 0, or
x ≠ 0, x ≠ Ꙥ/3, and x ≠ 2Ꙥ/3,
F(x) = f(x)/g(x) = (1 – 2sin 3x)/sin 3x > 0.
1. Solve f(x) = 1 – 2sin 3x = 0. This gives: sin 3x = 1/2 = sin 30⁰ = sin 150⁰.
a. sin 3x = sin 30⁰. This gives 3x = 30⁰ + k360⁰ → x = 10⁰ + k120⁰
b. sin 3x = sin 150⁰. This gives: 3x = 150⁰ + k360⁰ → x = 50⁰ + k120⁰
For k = 0, k = 1, k = 2, there are 6 end points at (10⁰), (50⁰), (130⁰), (170⁰), (250⁰), and
(290⁰).There are 6 arc lengths.
Find the sign status of f(x) inside the arc length (10⁰, 50⁰). Select as check point the
point (30⁰). We get: f(30⁰) = 1 – 2(sin 90⁰) = 1 – 2 < 0. Color it blue, and color the other
arc lengths (red, blue, red, blue, and red)
2. Solve g(x) = sin 3x = 0. On the unit circle sin 3x = 0 when:
a. sin 3x = sin 0. This gives: 3x = 0 + k360⁰ → x = 0 + k120⁰
b. sin 3x = 180⁰. This gives: 3x = 180⁰ + k360⁰ → x = 60⁰ + k120⁰
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c. sin 3x = 360⁰. This gives 3x = 360⁰ + k360⁰ → x = 120⁰ + k120⁰
For k = 0, k – 1, k = 2, there are 6 end points at: ((0⁰), (60⁰), (120⁰), (180⁰), (240⁰),
(240⁰), and (300⁰). There are 6 equal arc lengths.
Select point (30⁰) as check point. We get: f(30⁰) = sin 90⁰ = 1 > 0. Therefor,
g(x) = sin 3x > 0 in the interval: (0⁰, 60⁰). Color it red, and color the 5 other arc lengths
on the second concentric unit circle. Since the inequality (1) requires sin 3x > 0,
therefor, we can ignore the blue arc lengths where sin 3x < 0. See Figure 1.
By superimposing, we find that the solution set appears where f(x) and g(x) have the
same sign (red/red). They are the open intervals: (0⁰, 10⁰) and (50⁰, 60⁰), and (120⁰,
130⁰), and (170⁰, 180⁰)
Figure 1
Check.
Figure 2
F(x) = 1/sin 3x > 2
x = 5⁰. This gives: sin 3x = sin 15⁰ = 0.26 –→ F(x) = (1/0.26) = 3.85 > 2. Proved
x = 55⁰. This gives sin 3x = sin 165⁰ = 0.26 → F(x) = (1/0.26) = 3.85 > 2. Proved
x = 100⁰. This gives: sin 3x = sin 300⁰ = - 0.87 → F(x) = 1/-0.87 < 2. Proved
x = 125⁰.This gives: sin 3x = sin 375⁰ = 0.26 → F(x) = (1/0.26) = 3.85 > 2. Proved
x = 175⁰. This gives: sin 3x = sin 525⁰ = 0.26 → F(x) = (1/0.26) = 3.75 > 2. Proved
x = 235⁰. This gives: sin 3x = sin 705⁰ = -0.26 → F(x) = (1/ –0.26) < 2. Proved
x = 305⁰. This gives: sin 3x = sin 915⁰ = -0.26 → F(x) = (1/ –0.26> < 2. Proved
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Example 2. Solve
sin x + sin 3x + sin 5x > 0
(0, 2Ꙥ)
Solution. Using trig identity (sin a + sin b), we get
sin 3x + sin x + sin 5x = sin 3x + 2sin 3x.cos 2x = sin 3x.(1 + 2cos 2x) > 0
General form F(x) = f(x).g(x) = sin 3x.(1 + 2cos 2x)
1. First, solve f(x) = sin 3x = 0 = sin 0 = sin Ꙥ = sin 2Ꙥ
a. 3x = 0 + 2kꙤ → x = 0 + 2kꙤ/3
b. 3x = Ꙥ + 2kꙤ → x = Ꙥ/3 + 2kꙤ/3
c. 3x = 2Ꙥ + 2kꙤ → x = 2Ꙥ/3 + 2kꙤ/3
There are 6 end points at: (0), (Ꙥ/3), (2Ꙥ/3), (Ꙥ), (4Ꙥ/3), and (5Ꙥ/3). See Example 1.
There are 6 equal arc lengths. The sign status of f(x) = sin 3x is shown in Figure 1.
2. Solve g(x) = 2cos 2x + 1 = 0. This gives: cos 2x = - 1/2 = cos (2Ꙥ/3) = cos (4Ꙥ/3)
a. 2x = 2Ꙥ/3 + 2kꙤ → x = Ꙥ/3 + kꙤ.
b. 2x = 4Ꙥ/3 + 2kꙤ → x = 2Ꙥ/3 + kꙤ
For k = 0 and k =1, there are 4 end points at (Ꙥ/3), (2Ꙥ/3), (4Ꙥ/3), and (5Ꙥ/3).
There are 4 arc lengths.
Select test point (x = Ꙥ/2), we get g(Ꙥ/2) = 2cos Ꙥ + 1/2 = - 2 + 1/2 < 0.
Color the arc length (Ꙥ/3, 2Ꙥ/3) blue, and color the 3 other arc lengths.
By superimposing, we see that the solution set of F(x) > 0 is the open interval.
(0, Ꙥ/3), and (Ꙥ/3, 2Ꙥ/3), and (2Ꙥ/3, Ꙥ), or open combined interval (0, Ꙥ). Figure 2.
Check.
F(x) = sin 3x(2cos 2x + 1) > 0
x = Ꙥ/4 → F(x) = sin (3Ꙥ/4)(2cos Ꙥ/2 + 1) = (+) (+) > 0. Proved
x = Ꙥ/2 → F(Ꙥ/2) = sin (3Ꙥ/2)(2cos Ꙥ + 1) = (-1)(-2 + 1) = (-)(-) > 0. Proved
x = 140⁰ → F(140⁰) = sin 420⁰(2cos 280⁰ +1) = (0.87)(0.34 + 1) = (+)(+) > 0. Proved
x = (3Ꙥ/2) → F(x) = sin (9Ꙥ/2)(2cos 3Ꙥ + 1) = (1)(-2 + 1) = < 0. Proved
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Example 3. Solve
sin x + √3cos x < √2
Solution. Replace √3 by tan (Ꙥ/3), we get:
sin x.cos (Ꙥ/3) + sin (Ꙥ/3).cos x = √2.cos (Ꙥ/3) = √2(1/2) = √2/2
Applying the trig identity sin (a + b), we get:
sin (x + Ꙥ/3) = √2/2 = sin (Ꙥ/4) = sin (3Ꙥ/4). This gives 2 equations:
a. x + Ꙥ/3 = Ꙥ/4 + 2kꙤ → x = - Ꙥ/12 + 2kꙤ
b. x + Ꙥ/3 = 3Ꙥ/4 + 2kꙤ → x = 5Ꙥ/12 + 2kꙤ
For k = 0, there are 2 end points at (-Ꙥ/12), (5Ꙥ/12)
There are 2 arc lengths. To find the sign status of f(x), select point (0) as check point.
We have f(0) = sin x + √3cos x - √2 = 0 + √3 - √2 > 0. Therefor, f(x) is positive inside
the arc length (-Ꙥ/12, 5Ꙥ/12). Color it red and color blue the other arc length.
The solution for f(x) < 0 are the open blue interval: (5Ꙥ/12, 11Ꙥ/12). See Figure 3
Figure 3
Check.
Figure 4
f(x) = sin x + √3cos x - √2 < 0
x = 0. This gives: f(0) = 0 + √3 - √2 > 0. Proved
x = Ꙥ/2. This gives: f(Ꙥ/2) = 1 + 0 - √2 < 0. Proved
x = Ꙥ. This gives: f(Ꙥ) = 0 - √3 - √2 < 0. Proved
x = 3Ꙥ/2. This gives f(3Ꙥ/2) = - 1 + 0 - √2 < 0. Proved
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Example 4. Solve
tan³ x + tan² x – 3tan x > 3
(0, 2Ꙥ)
Solution. General form: F(x) = tan³ x + tan² x – 3tan x – 3 > 0
Condition: x ≠ Ꙥ/2 + kꙤ. Since: a – b + c – d = 0, one real root is (-1).
First solve F(x) = 0. Put (tan x + 1) in common factor, we get:
F(x) = (tan x + 1)(tan² x – 3) = (tan x + 1)(tan x - √3)(tan x + √3) = f(x)g(x)h(x) = 0
F(x) is undefined at x = Ꙥ/2, and x = 3Ꙥ/2.
1. Solve f(x) = 0. This gives tan x = -1 → x = - Ꙥ/4 + kꙤ.
There are 2 end points at: (-Ꙥ/4) and (3Ꙥ/4). There are 4 arc lengths for (0, 2Ꙥ).
Select point (0) as check point. We get: f(0) = 0 + 1 > 0. Therefor, f(x) > 0 inside
interval (- Ꙥ/4, Ꙥ/2). Color it red and color the other 3 intervals.
2. Solve g(x) = tan x - √3 = 0. This gives tan x = √3 → x = Ꙥ/3 + kꙤ.
There are 2 end points at: (Ꙥ/3), and (Ꙥ/2) where g(x) is undefined.
There are 4 arc lengths for (0, 2Ꙥ). Select point (Ꙥ) as check point.
We get: g(Ꙥ) = 0 – √3 < 0.
Therefor, g(x) < 0 inside the interval (-Ꙥ/2, Ꙥ/3). Color it blue and color other red.
3. Solve h(x) = tan x + √3 = 0. This gives: tan x = - √3 → x = - Ꙥ/3 + kꙤ
There are 2 end points at: (- Ꙥ/3) and (- Ꙥ/2) where h(x) is undefined.
There are 4 arc lengths for (0, 2Ꙥ)
Select point (0) as check point, we get h(0) = tan (0) + √3 > 0. Therefor, h(x) > 0 inside
the interval (- Ꙥ/3, Ꙥ/3).
By superimposing, we see that the solution set is the open interval (Ꙥ/3, Ꙥ/2) and
(2Ꙥ/3, 3Ꙥ/4) and (4Ꙥ/3, 3Ꙥ/2), and (5Ꙥ/3, 7Ꙥ/4). See Figure 4.
Check.
F(x) = (tan x + 1)(tan² x – 3) > 0
x = 75⁰. This gives: f(x) = 3.73 + 1 > 0, and g(x) = (13.93 – 3) > 0. Proved
x = 130⁰. This gives: f(x) = -1.19 + 1 < 0 and g(x) = 1.42 – 3 < 0. Proved
x = 180⁰. This gives: f(x) = 0 + 1 > 0, and g(x) = 0 – 3 < 0. Proved
x = 100⁰. This gives: f(x) = -5.67 + 1 < 0, and g(x) = 32.16 – 3 > 0. Proved.
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Example 5. Solve
2sin x – 5cos x > 5
Solution 1. General form: F(x) = 2sin x – 5cos x – 5 > 0
F(x) = 2sin x – 5(1+ cos x) > 0
First, solve F(x) = 0. The common period is (4Ꙥ = 720⁰)
Using 2 trig identities: sin a = 2sin a/2.cos a/2, and (1 + cos a) = 2cos² a/2, we get:
F(x) =4(sin x/2).(cos x/2) – 10cos² x/2 = f(x).g(x) = 2cos x/2.(2sin x/2 – 5cos x/2) = 0
1. Solve f(x) = cos x/2 = 0. This gives:
x/2 = ± Ꙥ/2 + 2kꙤ → x = ± Ꙥ + 4kꙤ
x = ± 3Ꙥ/2 + 2kꙤ →x = ±3Ꙥ + 4kꙤ
For the unit circle (0, 720⁰), there are 2 end points at: 180⁰, 540⁰. There are 2 equal
arc lengths. Select point (90⁰) as check point. We get: f(90⁰) = cos 45⁰ > 0. There for
f(x) > 0 inside the half circle interval (- 540⁰, 180⁰). Color it red and color the other half
circle blue.
2. Solve g(x) = 2sin x/2.- 5cos x/2) = 0. Divide by cos x/2 (with condition: cos x/2 ≠ 0,
or x ≠ 180⁰, and x ≠ 540⁰), we get: g(x) = 2tan x/2 – 5 = 0. This gives 2 solutions:
tan x/2 = 5/2 = tan 68⁰2 + k180⁰
tan x/2 = tan (68⁰2 + 180⁰) = tan 248⁰2 + k180⁰. This gives:
x = 136⁰4 + k360⁰, and
x = 496⁰4 + k360⁰
On the unit circle (0, 720⁰), there are 2 end points at: 136⁰4, and 496⁰4
There are 2 equal half circle intervals.
Select point (360⁰) as check point, we get g(180⁰) = 2(0) + 5 > 0. Therefor, g(x) > 0
inside the interval (136⁰2, 496⁰4). Color it red, and color the other blue. Figure 5.
By superimposing, we see that the solution set of F(x) > 0 is the open interval
(136⁰4, 180⁰) and (496⁰4, 540⁰)
Check.
F(x) = 2sin x – 5(1 + cos x) > 0
x = 170⁰ → sin x = 0.17, cos x = -0.98 → F(150) = 0.34 – 5(1 - 0.98) > 0. Proved
x – 520⁰ → sin x = 0.34; cos c = -0.93 → F(x) = 0.68 – 5(0.07) > 0. Proved
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Solution 2.
2sin x – 5cos x > 5 (1)
Divide both sides by 2. Call t the arc that tan t = 5/2 = tan (68⁰2), and cos t = 0.37.
The inequality (1) becomes:
sin x.cos t – sin t.cos x = 2.5.cos t = 0.928. Using trig identity sin (a – b), we get:
sin (x – t) = sin (x – 68⁰2) = sin (68.2) = sin (180 – 68.2) = sin 111⁰8. This gives
x – 68.2 = 68.2 + k360⁰ → x = 136⁰4 + k360⁰
x – 68.2 = 111.8 → x = 180⁰ + k360⁰
For k = 1, for the period (0, 720⁰), there are 4 arc lengths. Use check point (0), we get:
F(x) = 2sin x – 5cos – 5 → F(0) = 0 – 5 – 5 < 0.
Therefor, f(x) < 0 inside the blue interval (180⁰, 436⁰4) and ((540⁰, 136⁰4).
The solution set is the open red interval (136⁰4, 180⁰) and (436⁰4, 540⁰). It is the same
answer as the Solution 1. See Figure 6.
Figure 5
Check
Figure 6
F(x) = 2sin x – 5cos x – 5 < 0
x = 360⁰. This gives: F(360) = 0 – 5 – 5 < 0. Proved
x = 170⁰. This gives: F(170) = 0.35 + 4.92 – 5 > 0. Proved
x = 520⁰. This gives: F(520) = 0.68 – 5(-0.98) – 5 = 5.59 – 5 > 0. Proved.
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Example 6. Solve
1 + tan 2x > (1 – sin 2x)/cos² 2x
Condition: 2x ≠ Ꙥ/2 + kꙤ → x ≠ Ꙥ/4 + (kꙤ)/2
Solve: F(x) = 1 + tan 2x – (1 – sin 2x)/cos² 2x > 0
F(x) = cos² 2x + sin 2x.cos 2x – 1 + sin 2x > 0
Replace (cos² 2x - 1) by (- sin² 2x), the inequality becomes:
F(x) = sin 2x.(cos 2x – sin 2x + 1) = f(x).(g(x) > 0
1. Solve f(x) = sin 2x. The unit circle gives 2 solutions:
sin 2x = sin 0. This gives 2x = 0 + 2kꙤ → x = 0 + kꙤ
sin 2x = sin Ꙥ. This gives: 2x = Ꙥ + 2kꙤ → x = Ꙥ/2 + kꙤ
sin 2x = sin 2Ꙥ. This gives: 2x = 2Ꙥ + 2kꙤ → x = Ꙥ + kꙤ
General answer: x = k(Ꙥ/2)
For k = 0 and k = 1, there are 4 end points at: 0, Ꙥ/2, Ꙥ, and 3Ꙥ/2.
There are 4 equal arc lengths. Select point (Ꙥ/4) as check point, we get:
f(Ꙥ/4) = sin (Ꙥ/2) = 1 > 0.Then, F(x) > 0 inside the arc length (0, Ꙥ/2). Color it red and
color the 3 others.
2. Solve g(x) = (cos 2x – sin 2x) + 1) = -(sin 2x – cos 2x) + 1 = 0.
Using trig identity (sin a – cos a) = √2sin ( a – Ꙥ/4), we get:
g(x) = - sin (2x – Ꙥ/4) + 1 = 0 → sin (2x – Ꙥ/4) = √2/2 = sin Ꙥ/4 = sin (3Ꙥ/4).
This gives 2 solutions:
a. 2x – Ꙥ/4 = Ꙥ/4 + 2kꙤ → x = Ꙥ/4 + kꙤ (rejected since x must ≠ Ꙥ/4)
b. 2x – Ꙥ/4) = 3Ꙥ/4 + 2kꙤ → x = Ꙥ/2 + kꙤ
There are 2 end points at: Ꙥ/2 and 3Ꙥ/2. There are 2 infinities at: Ꙥ/4 and (3Ꙥ/4).
There are 4 arc lengths. Select 0 as check point, we get
g(0) = - sin (0 – Ꙥ/4) + 1 = - sin (-Ꙥ/4) + 1 = √2/2 + 1 > 0. There for, g(x) > 0 inside the
arc length (-3Ꙥ/2, Ꙥ/4). Color it red and color the others. See Figure 7.
By superimposing, the solution set of F(x) > 0 is the interval (0, Ꙥ/4) and (Ꙥ, 5Ꙥ/4).
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Check.
F(x) = sin 2x(cos 2x – sin 2x + 1) > 0
x = 20⁰ → F(20) = sin 40(cos 40 – sin 40 + 1) = 0.64(0.77 – 0.64 + 1) > 0. Proved
x = 200⁰ → F(200) = 0.64(0.77 – 0.64 + 1) > 0. Proved
x = 75⁰ → F(75) = 0.5(-0.87 – 0.5 + 1) < 0. Proved
x = 300⁰ → F(300) = - 0.87(-0.5 + 0.87 + 1) = (-)(+) < 0. Proved.
Figure 7
Figure 8
Note. Sometimes, it takes patience and some tricks in order to transform a complex
trig inequality into basic trig inequalities.
Example 7. Solve
F(x) = 2cos³ x + cos² x + sin x < 0
Solution. Solve F(x) = 0. Using trig identity 1 + cos 2a = 2 cos² a, we get
cos x(2cos² x) + cos 2x + sin x = 0
cos x(cos 2x + 1) + cos 2x + sin x = 0
cos 2x(cos x + 1) + cos x + sin x = 0
Replace cos 2x by (cos² x – sin² x)
(cos² x – sin² x)(cos x + 1) + cos x + sin x = 0
(cos x + sin x)(cos x – sin x)(cos x + 1) + (cos x + sin x) = 0
(cos x + sin x){(cos x – sin x)(cos x + 1) + 1} = 0
Factor (cos x + sin x)
(cos x + sin x)(cos² x + cos x - sin x cos x – sin x + 1} = 0
(cos x + sin x)[1 – sin² x + cos x – sin x cos x – sin x + 1] = 0 (cos² x = 1 – sin² x)
(cos x + sin x)[(1 – sin x)(1 + sin x) – cos x(1 - sin x) + (1 - sin x)] = 0
(cos x + sin x)[ (1 – sin x)(1 + sin x – cos x + 1)] = 0
Factor (1 – sin x)
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Finally, the inequality becomes:
F(x) = f(x).g(x).h(x) = (cos x + sin x)(1 – sin x)(sin x + cos x + 2) < 0
1. Solve f(x) = sin x + cos x = √2cos (x – Ꙥ/4) = 0. This gives
x – Ꙥ/4 = Ꙥ/2 + 2kꙤ → x = 3Ꙥ/4 + 2kꙤ
x – Ꙥ/4 = 3Ꙥ/2 + 2kꙤ → x = 7Ꙥ/4 + 2kꙤ
There are 2 end points at (3Ꙥ/4) and (7Ꙥ/4) and 2 arc lengths.
Take (Ꙥ) as check point, we get f(x) < 0 inside the arc length (3Ꙥ/3, 7Ꙥ/4).
2. Solve g(x) = 1 – sin x = 0 → sin x = 1 → x = Ꙥ/2. This function g(x) is always
positive regardless of x, except when x = Ꙥ/2.
3. Solve h(x) = sin x + cos x + 2 = 0. This function is always positive regardless of x.
Finally, the sign status of F(x) is the same as of f(x). The solution set of F(x) < 0 is the
interval (3Ꙥ/4, 7Ꙥ/4). Figure 8
Check
F(x) = 2cos³ x + cos² x + sin x < 0
(0, 2Ꙥ)
x = Ꙥ/2. This gives F(Ꙥ/2)) = 0 + 0 + 1 > 0. Proved
x = Ꙥ. This gives: F(Ꙥ) = -2 + 1 + 0 < 0. Proved
x = 3Ꙥ/2. This gives F(3Ꙥ/2) = 0 + 0 – 1 < 0. Proved
Example 8. Solve
cos 3x < ½
(0, 2Ꙥ)
Solution. General form f(x) = cos 3x – 1/2 < 0
First solve f(x) = cos 3x – 1/2 = 0 to find the end points. We have:
cos 3x = 1/2 = cos (± Ꙥ/3). This gives 2 equations:
a. cos 3x = cos Ꙥ/3 → 3x = Ꙥ/3 + 2kꙤ → x = Ꙥ/9 + (2kꙤ)/3
b. cos 3x = cos (-Ꙥ/3) → 3x = (5Ꙥ)/3 + 2kꙤ → x = (5Ꙥ)/9 + (2kꙤ)/3
For k = 0, k = 1, and k = 2, there are 6 end points at:
(Ꙥ/9), (5Ꙥ/9), (7Ꙥ/9), (11Ꙥ/9), (13Ꙥ/9), (17Ꙥ/9). There are 6 arc lengths.
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Select point (0) as check point. We get: F(0) = cos 0 – 1/2 = 1 – 1/2 > 0. The function
F(x) > 0 inside the arc length (17Ꙥ/9, Ꙥ/9 + 2Ꙥ). Color it red and color the 5 others.
The solution set is the blue interval (Ꙥ/9, 5Ꙥ/9) and (7Ꙥ/9, 11Ꙥ/9) and
(13Ꙥ/9, 17Ꙥ/9). See Figure 9.
Figure 9
Example 9. Solve
Figure 10
sin 3x + cos 3x > √2/2
(0, 2Ꙥ)
Solution. General form
F(x) = sin 3x + cos 3x - √2/2 > 0
Solve F(x) = 0. Applying the trig identity (sin a + cos a) = √2cos (a – 45⁰), we get:
sin 3x + cos 3x = √2.cos (3x – 45⁰) = √2/2
cos (3x – 45⁰) = 1/2 = cos (± 60⁰). There are 2 equations:
a. 3x – 45⁰ = 60⁰ + k360⁰ → 3x = 105⁰ + k360⁰ → x = 35⁰ + k120⁰
b. 3x – 45⁰ = 300⁰ + k360⁰ → 3x = 345⁰ + k360⁰ → x = 115⁰ + k120⁰
For k = 0, k =1 and k = 2, there are 6 end points at: 35⁰, 115⁰, 155⁰, 235⁰, 275⁰, 355⁰.
There are 6 arc lengths. Select point (0) as check point. We get: F(0) = 0 + 1 - √/2 > 0.
Therefor, F(x) > 0 inside the interval (355⁰, 395⁰). Color it red and color the other 5 arc
lengths.
The solution set of F(x) > 0 is the red interval (115⁰, 155⁰) and (235⁰, 275⁰) and
(355⁰, 395⁰). See Figure 10.
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Check.
F(x) = sin 3x + cos 3x - √2/2 > 0
x = 90⁰. This gives F(90) = sin (270) + cos (270) - √2/2 = - 1 + 0 - √2/2 < 0. Proved
x = 180⁰. This gives F(180) = sin (540) + cos (540) - √2/2 = 0 - 1 - √2/2 < 0. Proved
x = 270⁰. This gives F(270) = sin (810) + cos (810) - √2/2 = 1 – 0 - √2/2 > 0. Proved.
Example 10. Solve
sec x + csc x > 0
(0, 2Ꙥ)
Solution.
1/cos x + 1/ sin x > 0
(Condition: sin x ≠ 0, and cos x ≠ 0).
(sin x + cos x)/(1/2)sin x.cos x = 2(sin x+ cos x)/sin 2x > 0
F(x) = f(x)/g(x) = 2(sin x + cos x)/sin 2x = 2√2cos (x – Ꙥ/4)/sin 2x = f(x)/g(x)
1. Solve f(x) = cos (x – Ꙥ/4) = 0. This give 3 solutions:
a. x – Ꙥ/4 = 0 + 2kꙤ → x = Ꙥ4 + 2kꙤ
b. x – Ꙥ/4 = Ꙥ + 2kꙤ → x = 3Ꙥ/4 + 2kꙤ
c. x – Ꙥ/4 = 2Ꙥ + 2kꙤ → x = Ꙥ/4 + 2kꙤ
There are 2 end points at (Ꙥ/4) and (3Ꙥ/4). There are 2 arc lengths.
Select point (0) as check point. We get f(0) = cos Ꙥ/4 > 0.
Then, f(x) > 0 inside interval (3Ꙥ/4, 2Ꙥ + Ꙥ/4). Color it red and color the other half
circle blue
2. Solve g(x) = sin 2x = 0. This gives 3 solutions:
a. 2x = 0 + 2kꙤ → x = 0 + kꙤ
b. 2x = Ꙥ + 2kꙤ → x = Ꙥ/2 + kꙤ
c. x = 2Ꙥ → x = Ꙥ + kꙤ
For k = 0, and k = 1, there are 4 end points at (0), (Ꙥ/2), (Ꙥ), and (3Ꙥ/2). The function
g(x) > 0 inside the intervals (0, Ꙥ/2) and (Ꙥ, 3Ꙥ/2)
Figure the sign status of f(x) and g(x) on 2 concentric unit circles. By superimposing,
the solution set for F(x) > 0 is the intervals (0, Ꙥ/4) and (Ꙥ/2, 3Ꙥ/4) and (Ꙥ, 3Ꙥ/2).
See Figure 11.
Check
x = Ꙥ/8 → f(Ꙥ/8) = cos (Ꙥ/8 – Ꙥ/4) = cos (-Ꙥ/8) > 0. Proved
x = 100⁰ → f(100) = cos (100 – 45) = cos 55⁰ > 0. Proved
x = 200⁰ → f(200) = cos (200 – 45) = cos 155⁰ = - 0.91 < 0. Proved
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Figure 11
Example 11. Solve
Figure 12
sin (x – Ꙥ/4) + sin (x + Ꙥ/4) ≥ 1/√2
(0, 2Ꙥ)
Solution. Applying the trig identities:
(sin a + cos a) = √2 sin (a + Ꙥ/4) and (sin a – cos a) = √2sin (a – Ꙥ/4), we get:
1/√2 (sin x – cos x) + 1/√2 (sin x + cos x) = √2sin x = 1/√2 → sin x = 1/2
f(x) = sin x – 1/2 > 0
There are 2 end points at (Ꙥ/6) and (5Ꙥ/6) and 2 arc lengths. Select point (Ꙥ/2) as
check point, we get: f(Ꙥ/2) = sin (Ꙥ/2) – 1/2 = 1 – 1/2 > 0.
Therefor, the solution set for f(x) > 0 is the red interval (Ꙥ/6, 5Ꙥ/6). The 2 end points
are included in the solution set. See Figure 12.
Example 12. Solve
3tan x – cot x > 0
(0, 2Ꙥ)
Solution. 3(sin x/cos x) – cos x/sin x > 0
(3 sin² x – cos ² x)/sin x.cos x = (3sin² x – 1 + sin² x)/(sin 2x/2)
= 2(4sin² x – 1)/sin 2x = 2(2sin x – 1)(2sin x + 1)/sin 2x
F(x) = f(x).g(x)/h(x) = (2sin x – 1)(2sin x + 1)/sin 2x (Condition sin x ≠ 0 and cos x ≠ 0)
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1. Solve f(x) = 2sin x -1 = 0. This gives sin x = 1/2 = sin Ꙥ/6 = sin (5Ꙥ/6).
There are 2 end points at (Ꙥ/6) and (5Ꙥ/6) and 2 arc lengths.
Select point (Ꙥ/2) as check point. We get: f(Ꙥ/2) = 2(1) – 1 > 0. Therefor, f(x) > 0
inside the arc length (Ꙥ/6, 5Ꙥ/6). Color it red. And color the other arc length blue.
2. Solve g(x) = 2sin x + 1 = 0. This gives sin x = - 1/2 = sin (7Ꙥ/6) = sin (11Ꙥ/6).
There are 2 end points at (7Ꙥ/6) and (11Ꙥ/6) and 2 arc lengths.
Select point (3Ꙥ/2) as check point. We get: g(3Ꙥ/2) = 2(-1) – 1 < 0. Therefor, g(x) < 0
inside the arc length (7Ꙥ/6, and 11Ꙥ/6). Color it blue.
3. Solve h(x) = sin 2x = 0. This gives 4 end points at: (0), Ꙥ/2, Ꙥ, and 3Ꙥ/2.
There are 4 equal arc lengths. Select (Ꙥ/4) as check point. We get h(Ꙥ/4) = 1 > 0
h(x) > 0 inside the arc length (0, Ꙥ/2). Color it red and color the 3 others.
By superimposing, the solution set of F(x) > 0 is the interval ( Ꙥ/6, Ꙥ/2) and ( 5Ꙥ/6, Ꙥ) and
(11Ꙥ/6, 2Ꙥ). Note that F(x) is undefined when x = 0, x = Ꙥ/2, x = Ꙥ, and x = 3Ꙥ/2. Figure 13.
Check.
F(x) = 3tan x – cot x > 0
x = Ꙥ/3. This gives: F(Ꙥ/3) = 3√3 - √3/3 > 0. Proved
x = 5Ꙥ/4. This gives F(5Ꙥ/4) = tan (Ꙥ/4) – cot (Ꙥ/4) = 3 – 1 > 0. Proved.
x = 5Ꙥ/3. This gives: F(2Ꙥ/3) = - 3√3 - √3/3 < 0. Proved.
Figure 13
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Figure 14
Example 13. Solve
F(x) = cos 2x + sin x < 0
(0, 2Ꙥ)
Solution. Replace cos 2x by using trig identity: cos 2a = 1 – 2sin² a, then, solve F(x) = 0
F(x) = - 2sin² x + cos x + 1 = 0. Since a + b + c = 0, the 2 real roots are (1) and (-1/2).
F(x) = (sin x – 1)(- 2sin x – 1) = (-)(sin x – 1)(2sin x + 1) = 0.
Since (sin x – 1) is negative regardless of x, except when x = Ꙥ/2, therefor, the sign status of
F(x) is the same sign status as of f(x) = (2sin x + 1)
Solve f(x) = 2sin x + 1 = 0. This gives: sin x = - 1/2 = sin (7 Ꙥ/6) = sin (11 Ꙥ/6).
There are 2 end points at: (7Ꙥ/6) and (11Ꙥ/6) and 2 arc lengths. Select point (3 Ꙥ/2) as check
point, we get: f(3Ꙥ/2) = 2(-1) + 1 = -1 < 0. Therefor f(x) < 0 inside the interval (7 Ꙥ/6, 11 Ꙥ/6)
and that is the solution set of F(x) < 0. See Figure 14
Check.
F(x) = cos 2x + sin x < 0
x = 0. This gives: F(0) = cos 0 + sin 0 = 1 + 0 > 0. Proved
x = Ꙥ/2. This gives: F(Ꙥ/2) = cos Ꙥ + sin Ꙥ/2 = 0 + 1 > 0. Proved.
x = Ꙥ. This gives: F(Ꙥ) = cos 2Ꙥ + sin Ꙥ = 1 + 0 > 0. Proved
Note. After a few tries, if you are not able to transform a complex trig inequalities into basic
ones, then, you can use another method of transformation. This second transformation
method will be explained in the next math article.
(This math article is authored by Nghi H Nguyen, Dec 01, 2021)
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