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. Figure 1 Page 1 of 10 Figure 2 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 Solution 2csc² x < 2sec² x 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. By selecting the check point (0), we get: g(0) = 0 + 1 > 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) < 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) < 0 x = Ꙥ/2. This gives F(Ꙥ/2) = - (1 – 0)(1 + 0) = - 1 < 0. (Proved) x = 3Ꙥ/2. This gives F(3Ꙥ/2) = - (-1+ 0)(-1 +0) = - 1 < 0. (Proved) Page 2 of 10 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) Figure 3 Page 3 of 10 Figure 4 Exercises 4. Solve Solution. sin x - √3sin 3x + sin 5x < 0 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 + 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 > 0. Consequently, f(x) > 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 (± 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 < 0. So, g(x) < 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) < 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) < 0 x = 10⁰ → F(10) = sin 30(2cos 20 - √3) = 0.5(1.88 – 1.73) > 0 (Proved) x = 45⁰ --→ F(45) = sin 135.(2cos 90 - √3) = 0.71(0 – 1.73) < 0 (Proved) x = 90⁰ --→ F(90) = sin 270.(2cos 180 - √3) = - 1(-2 – 1.73) > 0 (Proved) x = 270⁰ --→ F(270) = sin 810.(cos 540 - √3) = 1(- 1 – 1.7) < 0 (Proved) Page 4 of 10 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 Page 5 of 10 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 Check. F(x) = 4sin x – 1 > 0 x = 60⁰ --→ F(x) = 4(1.73/2) – 1 > 0. Proved x = 170⁰ --→ F(x) = 4(0.17) – 1 = 0.68 – 1 < 0 Proved. x = 270⁰ --→ F(x) = - 3 – 1 = - 4 < 0. Proved Exercise 7. Solve sin 2x + √3.cos 2x > √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: Page 6 of 10 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 Solution tan 2x > 1 (Period Ꙥ) 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 Solve F(x) = tan 2x – 1 = 0. There are 2 solutions: Page 9 of 10 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 2Ꙥ) is located inside an arc length. 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. (This article was authored by Nghi H Nguyen – Feb. 10, 2021) Page 10 of 10