# Solving complex trig inequalities - Exercises Part II ```SOLVING COMPLEX TRIG INEQUALITIES – EXERCISES (Part II)
(Authored by Nghi H Nguyen – Feb 10, 2021)
Exercise 1. Solve
2sin x + 3cos x &lt; 2
Solution. Divide both side by 2, we get:
sin x + (3/2)cos x &lt; 1.
Call t the arc that tan t = 3/2. This gives t = 56⁰31 and cos t = 0.55.
sin x + (sin t/cos t).cos x &lt; 1
sin x.cos t + sin t.cos x &lt; cos t = 0.55 = sin (33⁰69)
Use the trig identity: sin (a + b) = sin a.cos.b + sin b.cos a.
sin (x + 56⁰31) = sin (33⁰69) → This leads to 2 solutions:
a. x + 56⁰31 = 33⁰69. This gives x = - 22⁰62, or x = 337⁰38 (co-terminal)
b. x + 56⁰31 = 180⁰ – 33⁰69 = 146⁰31. This gives x = 146⁰31 – 56.31 = 90⁰
There are 2 end points at x = 90⁰ and x = (337⁰38), and 2 arc lengths. To find the sign
status of F(x) = 2sin – 3cos x – 2&lt; 0, select the point (180⁰) as check point. We get:
F(180⁰) = 0 – 3 – 2 &lt; 0. Therefore, F(x) is negative in this interval (90⁰,337⁰38). Color it
blue and color the other arc length red.
The solution set of F(x) &lt; 0 is the open interval (90⁰, 337⁰38). See Figure 1.
Figure 1
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Figure 2
Check. F(x) = 2sin x + 3cos x – 2 &lt; 0
x = 270⁰. This give F(270⁰) = - 2 + 0 – 2 = - 4 &lt; 0. (Proved)
x = 0. This gives F(x) = 0 + 3 – 2 = 1 &gt; 0. (Proved)
Exercise 2. Solve
Solution
2csc&sup2; x &lt; 2sec&sup2; x
csc&sup2; x – sec&sup2; x &lt; 0
(csc x – sec x)(csc x + sec x) , 0
(1/sin x + 1/cos x)(1/sin x + 1/cos x) &lt; 0
(cos x – sin x)(cos x + sin x) &lt; 0
F(x) = f(x).g(x) = f’(x).g(x) = - (sin x – cos x)(sin x + cos x) &lt; 0, with f(x) = - f’(x).
1. Solve f’(x) = sin x – cos x = √2.sin (x – Ꙥ/4) = 0. There are 2 solutions
a. x – Ꙥ/4 = 0. This gives x = Ꙥ/4
b. x – Ꙥ/4 = Ꙥ. This gives x = 5Ꙥ/4
c. x – Ꙥ/4 = 2Ꙥ. This give x = 2Ꙥ + Ꙥ/4 = Ꙥ/4.
There are 2 end points at (Ꙥ/4) and (5Ꙥ/4), and 2 arc lengths. Select x = Ꙥ/2 as
check point. We get: f’(Ꙥ/2) = sin Ꙥ/2 – cos Ꙥ/2 = 1 – 0 = 1 &gt; 0. Then f’(x) &gt; 0 inside
the interval (Ꙥ/4, 5Ꙥ/4), and f(x) = -f’(x) is negative in this interval. Color it blue and
color the rest red.
2. Solve g(x) = sin x + cos x = √2 sin (x + Ꙥ/4) = 0. There are 3 solutions:
a. x + Ꙥ/4 = 0. This gives x = - Ꙥ/4 or x = 7Ꙥ/4 (co-terminal)
b. x + Ꙥ/4 = Ꙥ. This gives x = 3Ꙥ/4
c. x + Ꙥ/4 = 2Ꙥ. This gives x = 7Ꙥ/4.
By selecting the check point (0), we get: g(0) = 0 + 1 &gt; 0. Then, g(x) is positive inside
interval (-Ꙥ/4, 3Ꙥ/4). Color it red and color the rest blue.
By super imposing, the solution set of F(x) &lt; 0 are the 2 open intervals:
(Ꙥ/4, 3Ꙥ/4) and (5Ꙥ/4, 7Ꙥ/4)
Check.
F(x) = - (sin x – cos x)(sin x + cos x) &lt; 0
x = Ꙥ/2. This gives F(Ꙥ/2) = - (1 – 0)(1 + 0) = - 1 &lt; 0. (Proved)
x = 3Ꙥ/2. This gives F(3Ꙥ/2) = - (-1+ 0)(-1 +0) = - 1 &lt; 0. (Proved)
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Exercise 3. Solve
1 + 2cos&sup2; (x + Ꙥ/6) &lt; 3sin (Ꙥ/3 – x)
Solution. Use trig identity sin x = cos (Ꙥ/2 – x), we have:
3sin (Ꙥ/3 – x) = 3cos (Ꙥ/2 – Ꙥ/3 + x) = 3cos (x + Ꙥ/6). After transformation, we get:
F(x) = 2cos&sup2; (x + Ꙥ/6) – 3cos (x + Ꙥ/6) + 1 &lt; 0.
This is a quadratic equation in cos (x + Ꙥ/6). Since a + b + c = 0, the 2 real roots are
cos (x + Ꙥ/6) = 1 and cos (x + Ꙥ/6) = 1/2. After factoring we get:
F(x) = f(x).g(x) = [cos (x + Ꙥ/6) - 1].[2cos (x + Ꙥ/6) – 1) &lt; 0
1. f(x) = (cos (x + Ꙥ/6) – 1) is always negative regardless of x. So, the sign of F(x) is
the opposite sign of g(x)
2. Solve g(x) = 2cos (x + Ꙥ/6) – 1 = 0. That gives cos (x + Ꙥ/6) = 1/2 = cos (&plusmn; Ꙥ/3)
a. x + Ꙥ/6 = Ꙥ/3. This gives x = Ꙥ/3 – Ꙥ/6 = Ꙥ/6
b. x + Ꙥ/6 = - Ꙥ/3. This gives x = - Ꙥ/2.
There are 2 end points at (Ꙥ/6) and (-Ꙥ/2), and 2 arc lengths. Find the sign status of
g(x) by selecting the check point x = 0. We get: g(0) = 2cos (Ꙥ/6) – 1 = √3 – 1 &gt; 0.
Then, g(x) &gt; 0 and F(x) &lt; 0, opposite in sign to g(x), inside the interval (-Ꙥ/2, Ꙥ/6)
The solution set of F(x) &lt; 0 is the open interval (-Ꙥ/2, Ꙥ/6).
Check
F(x) = 2cos&sup2;(x + Ꙥ/6) – 3cos (x + Ꙥ/6) + 1 &lt; 0
x = 0 --→ F(0) = 2(3/4) – 3(√3/2) + 1 = 2.5 – 3(1.73) &lt; 0 (Proved)
x = Ꙥ --→ F(Ꙥ) = 2cos&sup2; (7Ꙥ/6) – 3cos (7Ꙥ/6) + 1 = 2(3/4) – 3(-√3/2) + 1 &gt; 0 (Proved)
Figure 3
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Figure 4
Exercises 4. Solve
Solution.
sin x - √3sin 3x + sin 5x &lt; 0
F(x) = sin x + sin 5x - √3sin 3x &lt; 0
Using trig identity (sin a + sin b) we get:
F(x) = 2sin 3x.cos 2x - √3sin 3x &lt; 0
F(x) = f(x).g(x) = sin 3x.(2cos 2x - √3) &lt; 0
1. Solve f(x) = sin 3x = 0. There are 3 solutions
a. 3x = 0 + 2kꙤ. This gives x = 0 + 2kꙤ/3
b. 3x = Ꙥ + 2kꙤ. This gives x = Ꙥ/3 + 2kꙤ/3
c. 3x = 2Ꙥ + 2kꙤ. This gives x = 2Ꙥ/3 + 2k
For k = 0, k = 1, and k = 2, there are 6 end points at: 0, Ꙥ/3, 2Ꙥ/3, Ꙥ, 4Ꙥ/3, and 5Ꙥ/3.
There are 6 equal arc lengths. Find the sign status of f(x) by selecting the check point
x = Ꙥ/6. We have f(Ꙥ/6) = sin (Ꙥ/2) = 1 &gt; 0. Consequently, f(x) &gt; 0 inside the interval
(0, Ꙥ/3). Color it red and color the 5 other arc lengths.
2. Solve g(x) = 2cos 2x - √3 = 0. This gives cos 2x = √3/2 = cos (&plusmn; Ꙥ/6)
a. cos 2x = cos Ꙥ/6 --→ x = Ꙥ/12 + kꙤ
b. cos 2x = cos (-Ꙥ/6) --→ x = - Ꙥ/12 + kꙤ, or x = 23Ꙥ/6 + kꙤ (co-terminal)
For k = 0, and k = 1, there are 4 end points at: (Ꙥ/12), (11Ꙥ/12), (13Ꙥ/12), and
(23Ꙥ/12). There are 4 arc lengths.
Find the sign status of g(x) by selecting the check point (Ꙥ/2). We get: g(Ꙥ/2) = 2cos
Ꙥ - √3 = -2 - √3 &lt; 0. So, g(x) &lt; 0 inside the interval (Ꙥ/12, 11Ꙥ/12). Color it blue and
color the other arc lengths. See Figure 4.
By superimposing, we see that the solution set of F(x) &lt; 0 are the 5 open intervals:
(Ꙥ/12, Ꙥ/3), (2Ꙥ/3, 11Ꙥ/12), (Ꙥ, 13Ꙥ/12), (4Ꙥ/3, 5Ꙥ/3), and (23Ꙥ/12, 2Ꙥ)
Fast check by calculator.
F(x) = sin 3x.(2cos 2x - √3) &lt; 0
x = 10⁰ → F(10) = sin 30(2cos 20 - √3) = 0.5(1.88 – 1.73) &gt; 0 (Proved)
x = 45⁰ --→ F(45) = sin 135.(2cos 90 - √3) = 0.71(0 – 1.73) &lt; 0 (Proved)
x = 90⁰ --→ F(90) = sin 270.(2cos 180 - √3) = - 1(-2 – 1.73) &gt; 0 (Proved)
x = 270⁰ --→ F(270) = sin 810.(cos 540 - √3) = 1(- 1 – 1.7) &lt; 0 (Proved)
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Exercise 5. Solve
F(x) = tan&sup2; x- (1 + √3).tan x + √3 &gt; 0
Solution. F(x) = tan&sup2; x – (1 + √3).tan x + √3 = 0
This is a quadratic equation in tan x. Since a + b + c = 0, so one real root is:
tan x = 1 and the other tan x = c/a = √3.
After factoring, we get:
F(x) = f(x).g(x) = (tan x -1)(tan x - √3) &gt; 0
(Period Ꙥ))
1. Solve f(x) = tan x - 1. This gives x = Ꙥ/4 . There is discontinuity at x = Ꙥ/2.
Find the sign status of f(x) by selecting (Ꙥ/6) as check point.
We get f(Ꙥ/6) = tan Ꙥ/6 – 1 = √3/3 – 1 &lt; 0. Then, f(x) &lt; 0 inside the interval (0, Ꙥ/4).
Color it blue and color the other intervals.
2. Solve g(x) = tan x - √3. This gives x = Ꙥ/3. There is discontinuity at x = Ꙥ/2.
Find the sign status of g(x) by selecting point (Ꙥ/4) as check pint. We get:
f(Ꙥ/4) tan Ꙥ/4 - √3 = 1 - √3 &lt; 0. Then, g(x) &lt; 0 inside the interval (0, Ꙥ/3). Color it blue
and color the other arc lengths.
By superimposing we see that the combined solution set of F(x) &gt; 0, are the 3 open
intervals: (0, Ꙥ/4), (Ꙥ/3, Ꙥ/2), and (Ꙥ/2, Ꙥ)
Fast check by calculator.
(tan x – 1)(tan x - √3) &gt; 0
x = 30⁰ --→ (tan 30 – 1)(tan 30 - √3) = (0,58 – 1)(0.58 - √3) &gt; 0. Proved
x = 50⁰ --→ (tan 50 – 1)(tan 50 - √3) = (1.19 – 1)(1.19 – 1.7) &lt; 0.Proved
x = 100 --→ (tan 100 – 1)(tan 100 - √3) = (-5.67 – 1)(-5.67 – 1.7) &gt; 0 Proved.
Figure 5
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Figure 6.
NOTE. This innovative method, sometimes, can simplify solving a complex trig
inequality into solving a basic trig inequality by one unit circle.
Exercise 6. Solve
sin 3x + sin x &gt; cos 2x – cos&sup2; x + 1
(0, 2Ꙥ)
Solution. Simplify the right side of the inequality by using trig identities.
RS = (1 + cos 2x) – cos&sup2; x = 2cos&sup2; x – cos&sup2; x = cos&sup2; x
Transform the left side by using the trig identity # 28: (sin a + sin b)
LS = sin 3x + sin x = 2.sin 2x.cos x = 4sin x.cos&sup2; x
After transformation, we get:
F(x) = 4sin x.cos&sup2; x – cos&sup2; x = cos&sup2; x.(4sin x – 1) &gt; 0
Since the term cos&sup2; x is always positive regardless of x, therefore, the sign status of
F(x) is the sign status of f(x) = (4sin x – 1).
Solve f(x) = 4sin x -1 = 0. This give sin x = 1/4. There are 2 solutions:
x = 14⁰48, and x = 180⁰ – 14⁰48 = 165⁰52.
Find the sign status of f(x) by selecting the check point (x = 90⁰). We get:
f(90) = 4 -1 = 3 &gt; 0. Consequently, f(x) &gt; 0 inside the interval (14⁰48, 165⁰52).
Consequently, the solution set of F(x) &gt; 0 is the open interval (14⁰48, 165⁰52). See
Figure 6
Check.
F(x) = 4sin x – 1 &gt; 0
x = 60⁰ --→ F(x) = 4(1.73/2) – 1 &gt; 0. Proved
x = 170⁰ --→ F(x) = 4(0.17) – 1 = 0.68 – 1 &lt; 0 Proved.
x = 270⁰ --→ F(x) = - 3 – 1 = - 4 &lt; 0. Proved
Exercise 7. Solve
sin 2x + √3.cos 2x &gt; √2 (1)
Solution. Call t the arc that: tan t = √3. We get t = Ꙥ/3 and cos t = cos Ꙥ/3 =1/2
The inequality (1) becomes: sin 2x.cos t + sin t.cos 2x = √2.cos t = √2/2
Using trig identity # 28 (sin a + sin b), we get:
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sin (2x + t) = sin (2x + Ꙥ/3) = √2/2 = sin Ꙥ.4. This gives 2 solutions:
a. 2x+ Ꙥ/3 = Ꙥ/4. This leads to: 2x = -Ꙥ/12 --→ x = - Ꙥ/24
b. 2x + Ꙥ/3 = Ꙥ – Ꙥ/4 = 3Ꙥ/4. This leads to: 2x = 5Ꙥ/12 --→ x = 5Ꙥ/24.
There are 2 end points at (5Ꙥ/24) and (- Ꙥ/24), and 2 arc lengths.
Find the sign status of F(x) = sin 2x + √3cos 2x - √2 &gt; 0.
Select the check point (x = 0). We get: F(0) = 0 + √3 - √2 &gt; 0. Therefore, F(x) &gt; 0
inside the interval (- Ꙥ/24, 5Ꙥ/24). Color it red and the other arc length blue.
The solution set of F(x) &gt; 0 is the red interval (- Ꙥ/24, 5Ꙥ/24)
Check.
F(x) = sin 2x + √3cos 2x - √2 &gt; 0.
x = Ꙥ/2. This gives F(Ꙥ/2) = 0 - √3 - √2 &lt; 0. Proved
x = 3Ꙥ/2. This give F(3Ꙥ/2) = 0 - √3 - √2 &lt; 0. Proved
Figure 7
Figure 8
Exercise 8. Solve
Solution
tan 2x &gt; 1
(Period Ꙥ)
F(x) = tan 2x – 1 &gt; 0
There are 2 discontinuities at:
2x = Ꙥ/2. This gives x = Ꙥ/4
2x = 3Ꙥ/2. This gives x = 3Ꙥ/4
Solve F(x) = tan 2x – 1 = 0. There are 2 solutions:
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a. tan 2x = 1 = tan Ꙥ/4 --→ 2x = Ꙥ/4 --→ x = Ꙥ/8
b tan 2x – 1 = tan (5Ꙥ/4) --→ 2x = 5Ꙥ/4 --→ x = 5Ꙥ/8.
There are 2 end points at (Ꙥ/8) and (5Ꙥ/8), and 5 arc lengths. See Figure 8.
The solution set of F(x) &gt; 0 are the 2 red open intervals: (Ꙥ/8, Ꙥ/4), and (5Ꙥ/8, 3Ꙥ/4)
Fast check by calculator.
F(x) = tan 2x – 1 &gt; 0
x = 10⁰ -→ F(10) = tan (20) – 1 = 0.36 – 1 &lt; 0 (Blue)
x = 40⁰ --→ F(40) = tan 80 – 1 = 5.67 -1 &gt; 0 (Red)
x = 90⁰ --→ F(90) = tan 180 – 1 = 0 – 1 &lt; 0 (Blue)
x = 130⁰ --→ F(130) = tan (260) – 1 = 5.67 – 1 &gt; 0 (Red)
x = 170⁰ --→ F(170) = tan (340) – 1 = -0.36 – 1 &lt; 0 (Blue)