SOLVING COMPLEX TRIG INEQUALITIES – EXERCISES (Part II) (Authored by Nghi H Nguyen – Feb 10, 2021) Exercise 1. Solve 2sin x + 3cos x < 2
Solution. Divide both side by 2, we get: sin x + (3/2)cos x < 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 < 1
sin x.cos t + sin t.cos x < 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< 0, select the point (180⁰) as check point. We get: F(180⁰) = 0 – 3 – 2 < 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) < 0 is the open interval (90⁰, 337⁰38). See Figure 1.
Check. F(x) = 2sin x + 3cos x – 2 < 0
x = 270⁰. This give F(270⁰) = - 2 + 0 – 2 = - 4 < 0. (Proved) x = 0. This gives F(x) = 0 + 3 – 2 = 1 > 0. (Proved)
Exercise 2. Solve 2csc² x < 2sec² x Solution csc² x – sec² x < 0 (csc x – sec x)(csc x + sec x) , 0
(1/sin x + 1/cos x)(1/sin x + 1/cos x) < 0 (cos x – sin x)(cos x + sin x) < 0
F(x) = f(x).g(x) = f’(x).g(x) = - (sin x – cos x)(sin x + cos x) < 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 > 0. Then f’(x) > 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.
Exercise 3. Solve 1 + 2cos² (x + Ꙥ/6) < 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) = 2cos² (x + Ꙥ/6) – 3cos (x + Ꙥ/6) + 1 < 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) < 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 (±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 > 0. Then, g(x) > 0 and F(x) < 0, opposite in sign to g(x), inside the interval (-Ꙥ/2, Ꙥ/6) The solution set of F(x) < 0 is the open interval (-Ꙥ/2, Ꙥ/6).
Check F(x) = 2cos²(x + Ꙥ/6) – 3cos (x + Ꙥ/6) + 1 < 0
x = 0 --→ F(0) = 2(3/4) – 3(√3/2) + 1 = 2.5 – 3(1.73) < 0 (Proved)
x = Ꙥ --→ F(Ꙥ) = 2cos² (7Ꙥ/6) – 3cos (7Ꙥ/6) + 1 = 2(3/4) – 3(-√3/2) + 1 > 0 (Proved)
Exercises 4. Solve sin x - √3sin 3x + sin 5x < 0 Solution. F(x) = sin x + sin 5x - √3sin 3x < 0 Using trig identity (sin a + sin b) we get:
F(x) = 2sin 3x.cos 2x - √3sin 3x < 0
F(x) = f(x).g(x) = sin 3x.(2cos 2x - √3) < 0 1. Solve f(x) = sin 3x = 0. There are 3 solutions a. 3x = 0 + 2k . This gives x = 0 + 2k /3Ꙥ Ꙥ b. 3x = + 2k . This gives x = /3 + 2k /3Ꙥ Ꙥ Ꙥ Ꙥ c. 3x = 2 + 2k . This gives x = 2 /3 + 2kꙤ Ꙥ Ꙥ
For k = 0, k = 1, and k = 2, there are 6 end points at: 0, /3, 2 /3, , 4 /3, and 5 /3.Ꙥ Ꙥ Ꙥ Ꙥ Ꙥ There are 6 equal arc lengths. Find the sign status of f(x) by selecting the check point x = /6. We have f( /6) = sin ( /2) = 1 > 0. Consequently, f(x) > 0 inside the interval Ꙥ Ꙥ Ꙥ (0, /3). Color it red and color the 5 other arc lengths.Ꙥ
2. Solve g(x) = 2cos 2x - √3 = 0. This gives cos 2x = √3/2 = cos (± /6)Ꙥ a. cos 2x = cos /6 --→ x = /12 + kꙤ Ꙥ Ꙥ
b. cos 2x = cos (- /6) --→ x = - /12 + k , or x = 23 /6 + k (co-terminal)Ꙥ Ꙥ Ꙥ Ꙥ Ꙥ For k = 0, and k = 1, there are 4 end points at: ( /12), (11 /12), (13 /12), and Ꙥ Ꙥ Ꙥ (23 /12). There are 4 arc lengths. Ꙥ
Find the sign status of g(x) by selecting the check point ( /2). We get: g( /2) = 2cos Ꙥ Ꙥ -
Ꙥ √3 = -2 - √3 < 0. So, g(x) < 0 inside the interval ( /12, 11 /12). Color it blue and Ꙥ Ꙥ color the other arc lengths. See Figure 4.
Exercise 5. Solve F(x) = tan² x- (1 + √3).tan x + √3 > 0 Solution. F(x) = tan² 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) > 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 < 0. Then, f(x) < 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 < 0. Then, g(x) < 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) > 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) > 0
x = 30⁰ --→ (tan 30 – 1)(tan 30 - √3) = (0,58 – 1)(0.58 - √3) > 0. Proved x = 50⁰ --→ (tan 50 – 1)(tan 50 - √3) = (1.19 – 1)(1.19 – 1.7) < 0.Proved x = 100 --→ (tan 100 – 1)(tan 100 - √3) = (-5.67 – 1)(-5.67 – 1.7) > 0 Proved.
Figure 5 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 > cos 2x – cos² x + 1 (0, 2Ꙥ) Solution. Simplify the right side of the inequality by using trig identities. RS = (1 + cos 2x) – cos² x = 2cos² x – cos² x = cos² 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.cos² x
After transformation, we get:
F(x) = 4sin x.cos² x – cos² x = cos² x.(4sin x – 1) > 0
Since the term cos² 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 > 0. Consequently, f(x) > 0 inside the interval (14⁰48, 165⁰52). Consequently, the solution set of F(x) > 0 is the open interval (14⁰48, 165⁰52). See Figure 6
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 > 0.
Select the check point (x = 0). We get: F(0) = 0 + √3 - √2 > 0. Therefore, F(x) > 0 inside the interval (- Ꙥ/24, 5Ꙥ/24). Color it red and the other arc length blue. The solution set of F(x) > 0 is the red interval (- Ꙥ/24, 5Ꙥ/24)
Check. F(x) = sin 2x + √3cos 2x - √2 > 0. x = Ꙥ/2. This gives F(Ꙥ/2) = 0 - √3 - √2 < 0. Proved x = 3Ꙥ/2. This give F(3Ꙥ/2) = 0 - √3 - √2 < 0. Proved
Figure 7 Figure 8
Exercise 8. Solve tan 2x > 1 (Period Ꙥ) Solution F(x) = tan 2x – 1 > 0
There are 2 discontinuities at: 2x = Ꙥ/2. This gives x = Ꙥ/4 2x = 3Ꙥ/2. This gives x = 3Ꙥ/4
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) > 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 > 0
x = 10⁰ -→ F(10) = tan (20) – 1 = 0.36 – 1 < 0 (Blue) x = 40⁰ --→ F(40) = tan 80 – 1 = 5.67 -1 > 0 (Red) x = 90⁰ --→ F(90) = tan 180 – 1 = 0 – 1 < 0 (Blue)
x = 130⁰ --→ F(130) = tan (260) – 1 = 5.67 – 1 > 0 (Red) x = 170⁰ --→ F(170) = tan (340) – 1 = -0.36 – 1 < 0 (Blue) ADVANTAGES OF NGHI NGUYEN METHOD
1. This new method can avoid the cut-offs at the 2 extremities of the sign chart. On the unit circle, these extremities joint together and show the periodic character of trig functions, whenever the origin (0 or 2 ) is located inside an arc length.Ꙥ
2. This method can often simplify the solving of a complex trig inequality into simple solving of a basic inequality.
3. The rules of end points and arc lengths make the solving process more convenient than the Sign Chart method.
