Solving Complex Trig inequalities – Method and Examples (Part 3)
(Authored by Nghi H Nguyen, Jan. 06, 2022)
GENERALITIES
After many tries, if you aren’t able to transform a complex trig inequality F(x) ≥ 0
(or ≤ 0) into basic ones, you may use another alternative transforming method
described below.
This second method tries to transform a complex trig inequality, containing 2 or a few
function variables, into an inequality containing only one function variable.
Transformation means include using trig identities and algebraic maneuvers. Next, we
solve this trig inequality as a basic one using the unit circle that is numbered in radians
or degrees.
Example of trig inequalities:
a. containing only one function variable:
F(x) = 2sin x + 3sin³ x > 0
b. Containing 2 function variables:
F(x) = 5sin 2x + 3cos x < 1
c. Containing a few function variables:
F(x) = tan x + cot x + 2cos x < 0
F(x) = cos² 2x + cos 2x – 3 < 0
F(x) = tan x + sec x ≤ √3
F(x) = sin 3x + cos 2x - cos² x + sin x > 0
TIPS FOR TRANSFORMATION
If the function F(x) doesn’t change when we replace:
a. x by (-x), choose cos x as function variable.
b. x by (Ꙥ – x), choose sin x as function variable.
c. x by (Ꙥ + x), choose tan x as function variable
d. If all the above operation works, select cos 2x as function variable.
e. If all the above operation do not work, select tan (x/2) as function variable.
Example 1. Solve
f(x) = 2sin² x - 5sin x – 3 > 0
Solution. For these types of trig inequalities, we solve f(x) = 0 as a quadratic equation.
Call sin x = t, we get a quadratic equation in t.
f(t) = 2t² – 5t – 3 = 0. There are 2 real roots: t1 = - 1/2, and t2 = 3 (rejected)
sin x = t = -1/2 = sin (7Ꙥ/6) = sin (11Ꙥ/6).
There are 2 end points at (7Ꙥ/6) and (11Ꙥ/6) and 2 arc lengths. Using point (0) as
check point, we get f(0) = -3 < 0.
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Therefor, the solution set is the red open interval (7Ꙥ/6, 11Ꙥ/6). See Figure 1.
Check.
f(x) = 2sin² x – 5sin x – 3 > 0
x = Ꙥ/2. This gives f(Ꙥ/2) = 2 – 5 – 3 < 0. Proved
x = (3Ꙥ)/2. This gives f(3Ꙥ/2) = 2 + 5 – 3 > 0. Proved
Figure 1
Example 2. Solve
Solution. General form:
Figure 2
sec²x > (√2 – 1) tan x > – √2 + 3
(0, 360⁰)
F(x) = sec² x - (√2 – 1)tan x + √2 – 3 > 0
Condition cos x ≠ 0 → x ≠ Ꙥ/2 and x ≠ 3Ꙥ/2.
First solve F(x) = 0. Using trig identity sec² x = 1/cos² x = 1 + tan² x, we get:
F(x) = 1 + tan² x - (√2 – 1) tan x – √2 + 3 = 0. After simplification:
F(t) = tan² x - (√2 – 1) tan x + √2 - 2 = 0. Call t = tan x, we get a quadratic equation in t:
F(t)= t² - (√2 – 1) t + √2 – 2 = 0. Since a + b + c = 0, one real root is (t1 = 1) and the
other is: t2 = c/a = √2 – 2 = - 0.59.
The real root t1 = 1 gives 2 end points at (45⁰) and (225⁰)
The real root t2 = - 0.59 gives 2 end points at (- 32⁰36) and (147⁰64)
Plot these 4 end points that divide the unit circle into 4 arc lengths.
Selecting point (0) as check point, we get f(0) = 0 – 0 + √2 – 2 < 0. Therefor, f(x) < 0
inside the arc length (- 32⁰36, 45⁰). Color it blue and color the 3 others. Figure 2
The solution set for F(x) > 0 are the 4 red intervals (45⁰, 90⁰) and (90⁰, 147⁰64) and
(225⁰, 270⁰) and (270⁰, - 32⁰36).
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Note. We can proceed another way. The function F(x) can be factored into:
F(x) = f(x).g(x) = (tan x – 1)(tan x + 0.59) > 0
Solve f(x) = 0. This gives 2 end points at 45⁰ and 225⁰. There are 4 arc lengths.
Solve g(x) =0. On the second concentric circle, there are 2 end points at (-32⁰36) and
(147⁰64) and 4 arc lengths. By superimposing, we find the same answer.
Check
f(t) = (t – 1)(t + 0.59) > 0
x = 180⁰ → t = 0 → f(t) = - 0.59 < 0. Proved
x = 60⁰ → t = √3 → f(√3) = (√3 – 1)(√3 + 0.59) = 0.73(2.21) > 0. Proved
x = 120⁰ → t = - √3 → f(-√3) = (-√3 – 1)(-√3 + 0.59) = (- 2.73)(- 1.14) > 0. Proved
Example 3. Solve
sin² x + sin^4 x < cos² x
(0, 2Ꙥ)
Solution. General form: F(x) = (sin² x – cos² x) – sin^4 x < 0
The inequality doesn’t change when x is replaced by (-x), (Ꙥ – x), (Ꙥ + x).
Select cos 2x = t as variable. Using trig identity: 1 – cos 2a = 2sin² a, and
cos²a – sin² a = cos 2x, we get:
F(t) = - t + (1/4)(1 – t)² = 0 = -4t + 1 – 2t + t² = t² – 6t + 1 = 0
This quadratic equation has 2 real roots: t1 = 3 + 2√2 (Rejected), and
t2 = 3 – 2√2 = 0.172 . We have:
a. t = cos 2x = 0.172 → 2x = ±80⁰10 + k360⁰ → x = ±40⁰05 + k180⁰
There are 4 end points at (40⁰05), (139⁰05), (220⁰05), (319⁰95) or (-40⁰05).
There are 4 arc lengths. Select point 0 as check point. We get F(t) = t² – 6t + 1 < 0
Therefor, F(t) < 0 inside the interval (-40⁰05, 40⁰05). See Figure 3.
Figure 3
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Figure 4
The solution set of F(x) < 0 are the 2 blue intervals (139⁰05, 220⁰05) and (-40⁰05, 40⁰05)
Check.
F(x) = sin² x + sin^4 x – cos² x < 0
x = 0 → F(0) = 0 + 0 – 1 < 0. Proved
x = 90⁰ → F(90) = 1 + 1 – 0 > 0. Proved
x = 180⁰ → F(180) = 0 + 0 – 1 < 0. Proved
x = 270⁰ → F(270) = 1 + 1 – 0 > 0. Proved
Example 4. Solve
sin^4 x + cos^4 x < 2cos 2x + 2sin² x
(0, 360⁰)
Solution. General form: F(x) = sin^4 x + cos^4 x – 2cos 2x – 2sin² x < 0
The equation doesn’t changed when x is replaced by (-x), (Ꙥ – x), and (Ꙥ + x).
Choose cos 2x = t as function variable.
Use trig identities: sin² a = (1- cos 2a)/2 and cos² a = (1 + cos 2a)/2 to transform.
General form: F(t) = (1- t²)/4 + (1 + t)²/4 - 2t - (1 – t) < 0
(1 – t)² + (1 + t)² - 4(t + 1) < 0
F(t) = t² – 2t – 1 < 0
Solve this quadratic equation F(t) = 0 for t = cos 2x.
The 2 real roots are: t = 1 ± √2
a. t = cos 2x = 1 + √2. Rejected because > 1
b. t = cos 2x = 1 - √2 = - 0.414 = cos ± 114⁰46
c. cos 2x = 114⁰46 + k360⁰--> x = 57⁰23 + k180⁰
d. cos 2x = - 114⁰46 + k360⁰ --> x = -57⁰23 + k180⁰
For k = 0, and k = 1, there are 4 end points at: (57⁰23), (122⁰77), (237⁰23), and
(- 57⁰23) or (302⁰77). There are 4 arc lengths.
Inside the arc length (-57⁰23, 57⁰23) select point 0 as check point. We get:
F(0) = 0 + 1 - 2 - 0 < 0.
Therefor, F(x) is negative inside this interval. Color it blue and color the 3 others. See
Figure 4.
The solution set of F(x) < 0 is the blue interval (- 57⁰23, 57⁰23), and (122⁰77, 227⁰23).
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Check.
F(x) = sin^4 x + cos^4 x – 2cos 2x – 2sin² x < 0
x = Ꙥ/2. This gives F(Ꙥ/2) = 0 +1 + 2 – 2 > 0. Proved
x = Ꙥ. This gives: F(Ꙥ) = 0 + 1 – 2 – 0 < 0. Proved
x = 3Ꙥ/2. This gives F(3Ꙥ/2) = 1 + 0 – (-2) - 2 = 1 > 0. Proved
Example 5. Solve
6sin x – 2cos³ x < 5sin 2xcos x
Solution. Divide both sides by cos³x, we get, with condition cos x ≠ 0:
F(x) = 6(sin x/cos x)(1/cos²x) – 2 - 10(sin x/cos²x) < 0
6t(1+ t²) – 2 - 10t = 0
6t + 6t³ – 2 – 10t = 0
6t³ – 4t – 2 = 0
[Call tan x = t,. Replace (1/cos²x) by (1 + t²)]
Solve this equation for t = tan x. Since a + b + c = 1, one real root is t = 1.
Factoring: F(t) = (t – 1)(3t² + 3t + 1) = 0
The quadratic function g(t) = (3t² + 3t + 1) is always positive. Its graph is a parabola
that stays above the x-axis.
Solve f(t) = tan x – 1. That gives: t = tan x = 1 = tan Ꙥ/4 = tan (5Ꙥ/4).
There are 2 end points at (Ꙥ/4) and (5Ꙥ/4) and 4 arc lengths.
Select point x = 0 as check point, we get F(t) = (-1)(1) < 0. Therefor, F(x) < 0 inside the
arc length (5Ꙥ/4, 9Ꙥ/4), and that is the solution set. See Figure 5.
Figure 5
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Figure 6
Check
F(x) = 6sin x – 2cos³ x - 5sin 2xcos x < 0
x = 0 → F(0) = 0 – 2 – 0 < 0. Prox = Ꙥ/2 → F(Ꙥ/2) = 6 – 0 - > 0. Proved
x = Ꙥ → F(Ꙥ) = 0 + 2 – 0 > 0. Proved.
x = (3Ꙥ/2) → F(3Ꙥ/2) = - 6 – 0 – 0 < 0. Proved
Example 6. Solve
2sin 2x + 3(cos x – sin x) > 3
(0, 2Ꙥ)>
Solution. General form: F(x) = 2sin 2x + 3(cos x – sin x) – 3 > 0
Call t = cos – sin x = √2cos (x + Ꙥ/4). (condition: -√2 ≤ t ≤ √2). We have:
t² = 1 – sin 2x → sin 2x = 1 – t². First, solve F(x) = 0 after replacement.
F(t) = 2(1 – t²) + 3t – 3 = 0
F(t) = -2t² + 3t - 1 = 0.
Solve this quadratic equation for t. The 2 real roots are t = 1 and t = 1/2
1. t = cos x – sin x = √2cos (x + Ꙥ/4) = 1
cos (x + Ꙥ/4) = √2/2 = cos ± 45⁰
a. x + 45⁰ = 45⁰ + k360⁰ → x = 0 + k360⁰
b. x + 45⁰ = - 45⁰ + k360⁰ → x = - 90⁰ + k360⁰
2. t = cos x – sin x = √2cos (x + 45⁰) = 1/2
cos (x + 45⁰) = 1/2√2 = 0.353 = cos ± 69⁰30
a. x + 45⁰ = 69⁰30 → x = 24⁰30 + k360⁰
b. x + 45⁰ = - 69⁰30 → x = - 114⁰30 + k360⁰
There are 4 end points at (0), (24⁰30), (- 114⁰30), and (-270⁰) and 4 arc lengths.
Select point (90⁰) as check point. We get F(90) = 0 + 4(- 1) – 4 = -8 < 0.
The solution set for F(x) > 0 is the interval (0, 24⁰30), and (- 114⁰30, 270⁰)
Check. F(x) = 2sin 2x + 3(cos x – sin x) – 3 > 0
x = 20⁰. This give F(20) = 1.28 + 4(0.94 – 0.34) – 3 > 0. Proved
x = 90⁰. This gives: F(90) = 0 + 4(0 – 1) – 3 < 0. Proved
x = 250⁰. This gives: F(250) = 2(0.64) + 3(- 0.34 + 0.94) - 3 = 3.28 – 3 > 0. Proved
x = 315⁰. This gives: F(315) = 2(-1) + 3(0.71 – 0.71) – 3 < 0. Proved
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Example 7. Solve
tan x + 2 tan²x < cot x + 2
(0, 180⁰)
Solution. Select tan x = t since it doesn’t change when x is replaced by (x + Ꙥ).
General form: F(t) = t + 2t² – 1/t – 2 < 0
Condition t ≠ 0 --> x ≠ 0, x ≠ kꙤ, and t ≠ Ꙥ/2, and t ≠ 3Ꙥ/2)
Solve F(t) = 0. We get: F(t) = t²(1 + 2t) – (1 + 2t) = (1 + 2t) (t – 1)(t + 1)= f(t).g(t).h(t) = 0
a. Solve f(t) = (1 + 2t) = 0. That gives t = -1/2 = tan 153⁰43 --> x = 153⁰43. There are 1
end points at (153⁰43). Select point (x = 45⁰) as check point. We get f(45) = 1 + (2√3)/3
> 0. Therefor f(t) > 0 inside the arc length (0, 90⁰). By the same method, f(t) < 0 inside
the arc length (90⁰, 153⁰43), and f(t) > 0 inside the arc length (153⁰43, 180⁰)
b. Solve g(t) = t – 1. This gives t = tan x = 1 = tan 45⁰ → x = 45⁰
There are 1 end points at (45⁰), and 3 arc lengths.
Select point x = 30⁰ as check point we get g(30) = √3/3 + 6/9 < √3 + 2 = 1.24 < 3.73.
Therefor, g(t) is negative [g(t) < 0] inside the interval (0, 45⁰). Color it blue. By the
same check point method, we find that g(t) < 0 inside the arc length (90⁰, 180⁰)
c. Solve h(t) = t + 1 = 0. That gives one end point at (135⁰) and 3 arc lengths.
h(t) > 0, inside the arc length (0⁰, 90⁰) and (135⁰, 180⁰).
h(t) < 0 inside the arc length (90⁰, 135⁰)
By superimposing, the solution set of F(x) < 0 is the interval (0⁰, 45⁰) and (90⁰, 135⁰),
and (153⁰43, 180⁰) See Figure 7.
Figure 7
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Figure 8
Check.
F(t) = (1 + 2t)(t² – 1) < 0
x = 30⁰. This gives: F(t) = (1 + 2tan 30⁰)(tan² 30⁰ – 1) = (+)(-) < 0. Proved
x = 60⁰. This gives F(t) = (1 + 2tan 60)(tan² 60) = (+)(+) > 0. Proved
x = 120⁰. This gives F(t) = (-)(+) < 0. Proved
x = 150⁰. This gives F(t) = (-)(-) > 0. Proved
x = 170. This gives F(t) = (=)(-) < 0. Proved.
Example 8. Solve
cos 2x – 4cos x + 2 < 7sin² x - 10
(0, 360⁰)
Solution. General form: F(x) = cos 2x – 4cos x +2 - 7sin² x + 10 < 0
Choose cos x = t as function variable using trig identities:
cos 2x = 1 – 2t²; sin² x = 2 - t²
F(t) = (1 - 2t²) – 4t + 2 - 7(2 – t²) + 10 = 0
F(t) = 5t² – 4t – 1 = 0
Solve this quadratic equation for t. Since a + b + c = 0, one real root is (1) and the
other is (-1/5). We get the inequality by factoring:
F(t) = f(x).g(x) = (cos x – 1)(5cos x + 1) < 0
The function f(x) = (cos x – 1) is always negative regardless of x, except when x = 0 or
x = 2Ꙥ. Therefor, the sign status of F(x) is the opposite sign status of g(x) = 5cos x + 1
Solve g(x) = 5cos x + 1 = 0. This gives cos x = - 1/5 = cos ± 101.54.
There are 2 end points at (101⁰54) and (- 101⁰54), and 2 arc lengths. Select point (0)
as check point. We get g(0) = 5 + 1 = 6 > 0, and g(x) > 0 inside the interval
(-101⁰54, 101⁰54) that includes point (0). See Figure 8
Therefor, by contrary, the solution set of F(x) < 0 is the blue interval (-101⁰54, 101⁰54).
Check
F(t) = (-)(5cos x + 1) < 0
x = 90⁰ → cos x = 0 → F(cos x) = (-)(1) < 0. Proved
x = 180⁰ → cos x = - 1 → F(180) = (-)(-5 + 1) > 0. Proved
x = 270 → cos x = 0 → F(270) = (-)(1) < 0. Proved
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Example 9. Solve
sin x + 2cos x < 2cot (x/2) – 1
(0, 360⁰)
Select t = tan (x/2) and transform the inequality. Condition sin (x/2) ≠ 0 → x ≠ kꙤ
2t/(1 + t²) + 2(1 – t²)/(1 + t²) = (2 - t)/t
2t² + 2t – 2t³ = 2 + 2t² – t – t³
F(t) = t³ – 3t + 2 = 0
Since a + b + c = 0, one real toot is (1). Factoring:
F(t) = (t – 1)(t² + t – 2) = (t – 1)(t – 1) (t + 2).
The function f(t) = t – 1 is always negative regardless of x, except when t = 1 -->
when x /2 = 45⁰ → x = 90⁰.
Therefor, the sign status of (Ft) = (-)(-)g(t) is the same sign status as of g(t) = (t + 2)
F(t) = f(x).f(x).g(x) = (-)(-)(tan x/2 + 2) < 0
Solve g(x) = tan x/2 + 2 = 0
tan x/2 = - 2 = tan (- 63⁰44) + k 180⁰ = tan (- 63⁰44 + 180⁰) = tan 116⁰57.
Reminder: - 63⁰44 = 296⁰57 (co-terminal)
On the unit circle, g(x) = tan x/2 + 2 < 0 when x/2 varies inside the 2 arc lengths
(90⁰, 116⁰57) and (270⁰, 296⁰57). We get the same interval where g(x) < 0:
a. 90⁰ < x/2 < 116⁰57 --→ 180⁰ < x < 233⁰14
b. 270⁰ < x/2 < 296.57 --→ 540⁰ < x < 593.14, or 180⁰ < x < 233⁰14
Therefor, the solution set for F(x) = (-)(-)(tan x/2 + 2) < 0 is the interval (180⁰, 233⁰14).
Figure 9
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Figure 10
Check.
F(t) = (+) (tan x/2 + 2) < 0
x = 80⁰ → F(t) = tan 40⁰ + 2 > 0. Proved
x = 100⁰ → F(t) = tan 50⁰ + 2 = 1.19 + 2 > 0. Proved
x = 200⁰ → F(t) = tan 100⁰ + 2 = -5.67 + 2 < 0. Proved
x = 300⁰ → F(t) = tan 150⁰ + 2 = - 0.58 + 2 > 0. Proved
Example 10.
Solve
cos 2x + sin² x > 2cos x + 3
(0, 360⁰)
Solution. The inequality doesn’t change if x is replaced by (-x). Select cos x = t as
variable.
F(t) = (2t² – 1) + 5(1 – t²) – 2t – 3 = 0
F(t) = -3t² – 2t + 1 = 0
F(t) = 3t² + 2t – 1 = 0
Since a – b + c = 0, this quadratic equation has 2 real roots: t = -1 and t = 1/3.
Factoring: F(t) = ft)(g(t) = (t + 1)(3t – 1) > 0
The function f(t) = (t + 1) is always positive regardless of x, except when cos x = -1 -->
x = 180⁰).
Therefor the sign status of F(t) is the same as of g(t) = (3t – 1)
Solve g(t) = 3t – 1 = 3cos x – 1 = 0
This gives 3cos x = 1 –> cos x = 1/3 = cos ± 70⁰53.
There are 2 end points at (± 70⁰53) and 2 arc lengths. Select point 0 as check point.
We get g(t) = 3 – 1 > 0.
The solution set of F(x) > 0 is the red interval (- 70⁰53, 70⁰53). See Figure 10.
Check.
F(x) = (+)(3cos x – 1) > 0
x = 90⁰. This give cos x = 0 → F(0) = - 1< 0. Proved
x = 180⁰. This gives cos x = -1 → F(180) = -3 – 1 < 0. Proved
x = - 60⁰. This gives cos x =1/2 → F(-60) = 3/2 – 1 > 0. Proved
(This math article is authored by Nghi H Nguyen, on Jan. 06, 2022)
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