Trigonometric Inequality

An inequality of the standard form R(x) > 0 (or < 0) that consists of 1 or a few trigonometric functions of the variable arc x is a trigonometric inequality. Finding a solution to the inequality means determining the values of the variable arc x whose trigonometric functions make the inequality true. The solution sets of trigonometric inequalities are expressed in intervals.

The trigonometric unit circle is defined as a circle with radius one unit with origin O. The rotation of the variable arc in the counterclockwise direction on the trigonometric unit circle explains 4 common trigonometric functions of the arc namely,

  • cos x
  • sin x
  • tan x
  • cot x

Trigonometric Inequality Examples

The trigonometric unit circle can be used as proof in solving trigonometric inequalities.

The least multiple of all the periods of the trigonometric functions in the inequality is the common period of trigonometric inequality.

The 4 basic types of trigonometric inequalities are

  • sin x < a (or > a)
  • cos x < a (or > a)
  • tan x < a (or > a)
  • cot x < a (or > a)

where a is a given number.

The steps to find the solution for trigonometric inequality are as follows:

1] The inequality is converted into a trigonometric equation by using the equality sign in the place of the inequality sign.

2] The trigonometric equation is solved and solutions are obtained as angle values in the interval [0, 2π].

3] The positive angle greater than π is converted to the equivalent negative value for the fact that the repeated basic interval may lie on the negative side of the origin.

4] The base interval is developed between two values.

5] If the function asymptotes within the interval are developed, then the angle value at which the function asymptotes, limits the value of the basic interval.

6] Finally, the solution is generalized.

Solved Problems On Trigonometric Inequality

Example 1: Solve trigonometric inequality given by sin x ≥ 1 / 2.

Solution:

The solution of the corresponding equal equation is obtained as :

sin x = 1 / 2 = sin π / 6

⇒ x = π / 6

The sine function is positive in the first and second quarter. Hence, the second angle between “0” and “2π” is :

⇒ x = π − θ = π − π / 6 = 5π / 6

Both angles are less than “π”. Thus, we do not need to convert the angle into an equivalent negative angle. Further, the sine curve is defined for all values of “x”. The base interval, therefore, is π / 6 ≤ x ≤ 5π / 6

The periodicity of the sine function is “2π”. Hence, we add “2nπ” on either side of the base interval :

2nπ + [π / 6] ≤ x ≤ 2nπ + [5π / 6], n ∈ Z

Example 2: Solve trigonometric inequality given by sinx > cosx.

Solution:

In order to solve this inequality, it is required to convert it in terms of inequality of a single trigonometric function.

sinx > cosx

⇒ sinx − cosx > 0

⇒ sin x * cos π / 4 − cos x * sin π / 4 > 0

⇒ sin (x − π / 4) > 0

Let y = x − π/4.

Then,

sin y > 0

Thus, we see that problem finally reduces to solving trigonometric sine inequality. The solution of the corresponding equality is obtained as :

⇒ sin y = 0 = sin 0 ⇒ y = 0

The second angle between “0” and “2π” is “π”. The base interval, therefore, is : 0<y<π

The periodicity of the sine function is “2π”. Hence, we add “2nπ” on either side of the base interval :

2nπ < y < 2nπ + π, n ∈ Z

Now substituting for y = x − π/4, we have :

2nπ < [x − π / 4] < 2nπ + π, n ∈ Z

⇒ 2nπ + π / 4 < x < 2nπ + 5π / 4, n ∈ Z

Example 3: If the domain of a function, “f(x)”, is [0,1], then find the domain of the function given by f (2 sinx − 1).

Solution:

The domain of the function is given here. We need to find the domain when the argument (input) to the function is a trigonometric expression. The given domain is : 0≤x≤1

Changing the argument of the function, the domain becomes :

0 ≤ 2 sinx − 1 ≤ 1 ⇒ 1 ≤ 2 sinx ≤ 2 ⇒ ½ ≤ sin x ≤ 1

However, the range of sinx is [-1, 1]. It means that the above interval is equivalent to a trigonometric inequality given by sin x ≥ 1 / 2

The sine function is positive in the first and second quadrant. Two values of “x” between “0” and “2π” are :

π / 6, π − π / 6

⇒ π / 6, 5π / 6

The value of “x” satisfying above equation :

2nπ + π/6 < = x < = 2nπ + 5π/6, n ∈ Z

Hence, the required domain is :

[2nπ + π / 6, 2nπ + 5π / 6], n ∈ Z

Example 4: Solve the inequality tan x < 1.

Solution:

Hints: Here, the corresponding trigonometric equation is: tanx = 1 = tan π / 4

The other solution in [0, 2π] is x = π + π / 4 = 5π / 4

Corresponding negative angle: y = 5π / 4 − 2π = −3π / 4

However, the tangent function asymptotes at – π/2. Hence, the basic interval is :(−π / 2, π / 4)

Further, the tangent function has a period of π. The general solution is :

nπ − π / 2 < x < nπ + π / 4; n ∈ Z

Example 5: If 12cot2θ31cosec θ+32=012{{\cot }^{2}}\theta -31\,\text{cosec }\theta +\text{32}=\text{0}, then find the value of sinθ\sin \theta.

Solution:

12cot2θ31cosecθ+32=012(cosec2θ1)31cosec θ+32=012cosec2θ31cosec θ+20=012cosec2θ16cosec θ15cosecθ+20=0(4cosecθ5)(3cosecθ4)=0cosecθ=54,43; sinθ=45,34.12{{\cot }^{2}}\theta -31\cos ec\theta +32=0\\ 12(\text{cos}\text{e}{{\text{c}}^{2}}\theta -1)-31\text{cos}\text{ec }\theta +\text{32}=\text{0}\\ 12\text{cos}\text{e}{{\text{c}}^{2}}\theta -31\,\text{cos}\text{ec }\theta +\text{20}=\text{0}\\ 12\,\text{cos}\text{e}{{\text{c}}^{2}}\theta -16\,\,\text{cos}\text{ec }\theta -15\text{cos}\text{ec}\theta +20=\text{0}\\ (4\cos \text{ec}\theta -5)(3\cos \text{ec}\theta -4)=0\\ \text{cos}\text{ec}\theta =\frac{5}{4},\frac{4}{3}; \ \\ \sin \theta =\frac{4}{5},\frac{3}{4}.

Example 6:  The only value of x for which 2sinx+2cosx>21(1/2){{2}^{\sin x}}+{{2}^{\cos x}}> {{2}^{1-(1/\sqrt{2})}} holds, is 

A)5π4B)3π4C)π2A) \frac{5\pi }{4}\\ B) \frac{3\pi }{4}\\ C) \frac{\pi }{2}

D) All values of x

Solution: 

Since A.M. ≥ G.M.

12(2sinx+2cosx)2sinx.2cosx2sinx+2cosx2.2sinx+cosx22sinx+2cosx21+sinx+cosx2\frac{1}{2}({{2}^{\sin x}}+{{2}^{\cos x}})\ge \sqrt{{{2}^{\sin x}}{{.2}^{\cos x}}}\\ \Rightarrow {{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2.2}^{\frac{\sin x+\cos x}{2}}}\\ \Rightarrow {{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{1+\frac{\sin x+\cos x}{2}}} and we know that 

sinx+cosx2 Therefore 2sinx+2cosx>21(1/2), for x=5π4\sin x+\cos x\ge -\sqrt{2} \\ \text \ Therefore \ {{2}^{\sin x}}+{{2}^{\cos x}}>{{2}^{1-(1/\sqrt{2})}}, \text \ for \ x=\frac{5\pi }{4}

Example 7: Common roots of the equations 2sin2x+sin22x=22{{\sin }^{2}}x+{{\sin }^{2}}2x=2 and sin2x+cos2x=tanx,\sin 2x+\cos 2x=\tan x, are

A)x=(2n1)π2B)x=(2n+1)π4C)x=(2n+1)π3A) x=(2n-1)\frac{\pi }{2}\\ B) x=(2n+1)\frac{\pi }{4} \\C) x=(2n+1)\frac{\pi }{3}

D) None of these

Solution:

2sin2x+sin22x=2(i)sin2x+cos2x=tanx(ii)2{{\sin }^{2}}x+{{\sin }^{2}}2x=2 \rightarrow (i) \\ \sin 2x+\cos 2x=\tan x \rightarrow (ii)

Solving (i), 

sin22x=2cos2x2cos2xcos2x=0x=(2n+1)π2 or x=(2n+1)π4{{\sin }^{2}}2x=2{{\cos }^{2}}x\\ 2{{\cos }^{2}}x\cos 2x=0\\ x=(2n+1)\frac{\pi }{2}\text{ or }x=(2n+1)\frac{\pi }{4}

Common roots are (2n±1)π4(2n\pm 1)\frac{\pi }{4}

Solving (ii),

2tanx+1tan2x1+tan2x=tanxtan3x+tan2xtanx1=0(tan2x1)(tanx+1)=0x=mπ±π4\frac{2\tan x+1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\tan x\\ \Rightarrow {{\tan }^{3}}x+{{\tan }^{2}}x-\tan x-1=0\\ \Rightarrow ({{\tan }^{2}}x-1)\,(\tan x+1)=0\\ \Rightarrow x=m\pi \pm \frac{\pi }{4}

Example 8: If sinθ=3cosθ,π<θ<0,\sin \theta =\sqrt{3}\cos \theta ,-\pi <\theta <0, then find θ\theta.

Solution:

tanθ=3=tanπ3θ=nπ+π3\tan \theta =\sqrt{3}=\tan \frac{\pi }{3}\Rightarrow \theta =n\pi +\frac{\pi }{3}

For π<θ<0-\pi <\theta <0 

Put n = −1, we get 

θ=π+π3=2π3or 4π6.\theta =-\pi +\frac{\pi }{3}=\frac{-2\pi }{3}\text{or }\frac{-4\pi }{6}.

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