Trigonometric Equation 1€¦ · E The general solution should be given unless the solution is...

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Trigonometric Equation 1

1. TRIGONOMETRIC EQUATIONS An equation involving one or more trigonometrical ratios of unknown angle is called trigonometric equation e.g. cos 2 x – 4 sinx = 1. It is to be noted that a trigonometrical identity is satisfied for every value of the unknown angle whereas, trigonometric equation is satisfied only for some values (finite or infinite in number) of unknown angle. e.g. sin 2 x + cos 2 x = 1 is a trigonometrical identity as it is satisfied for every value of x∈R.

2. SOLUTION OFATRIGONOMETRIC EQUATION A value of the unknown angle which satisfies the given equation is called a solution of the equation

e.g. 6 / π = θ is a solution of sinθ = 2 1 .

3. GENERAL SOLUTION Since trigonometrical functions are periodic functions, solutions of trigonometric equations can be generalized with the help of the periodicity of the trigonometrical functions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution. We use the following formulae for solving the trigonometric equations: ) I n ( ∈

Ø sinθ = sinα ⇒ θ = nπ + (– 1) n α, where α π π

∈ − L N M O Q P ∈ 2 2 , , n I .

Ø cosθ = cosα ⇒ θ = 2nπ ± α, where α π ∈ ∈ 0, , n I .

Ø tanθ = tanα ⇒ θ = nπ + α, where α π π

∈ − F H G I K J ∈ 2 2 , , n I .

Ø sin 2 θ = sin 2 α ⇒ θ = nπ ± α, where α π ∈ L N M O Q P ∈ 0 2 , ,n I .

Ø cos 2 θ = cos 2 α ⇒θ = nπ ± α, where α π ∈ L N M O Q P ∈ 0 2 , ,n I .

Ø tan 2 θ = tan 2 α ⇒ θ = nπ ± α, where α π ∈ L N M I K J ∈ 0 2 , ,n I .

Note: (For these type of equations students must use these solution to get correct answers)

Ø sin θ = 0 ⇒ θ = nπ,

Ø cos θ = 0 ⇒ θ = (2n + 1) 2 π ,

Ø tan θ = 0 ⇒ θ = nπ,

Ø sinθ = 1 ⇒ θ = (4n + 1) 2 π

Ø sinθ = –1 ⇒ θ = (4n – 1) 2 π


2 Trigonometric Equation

Ø cosθ = 1 ⇒ θ = 2nπ

Ø cosθ = –1 ⇒ θ = (2n + 1)π

Ø sinθ = sinα and cosθ = cosα ⇒ θ = 2nπ + α

Note:

E Everywhere in this chapter n is taken as an integer, if not stated otherwise.

E The general solution should be given unless the solution is required in a specified interval or range.

DRILL EXERCISE 1

Solve the following trigonometric equation

1. tan θ + cot θ = –2 2. sin θ = 2 1 and cos θ =

2 3

−

3. cos 2 θ – sin θ – 4 1 = 0 4. sin sin sin θ θ θ + = 5 3 , where 0 ≤ ≤ θ π

5. sin cos 2 1 4

θ θ − = , where 0 2 ≤ ≤ θ π 6. tan 4 θ – 2 tan 2 θ – 3 = 0

4. TYPES OFTRIGONOMETRIC EQUATIONS :

(a) Solution of equations by factorising.Consider the equation ; (2 sin x cos x) (1 + cos x) = sin 2 x.

(b) Solutions of equations reducible to quadratic equations. Consider the equation ; 3 cos 2 x 10 cos x + 3 = 0

(c) Solving equations by introducing an Auxilliary argument. Consider the equation ;

sin cos x x + = 2 and 3 2 cos sin x x + = .

(d) Solving equations by Transforming a sum of Trigonometric functions into a product. Consider the example ; sin 5 x + sin 2 x sin 4x = 0

(e) Solving equations by transforming a product of trigonometric functions into a sum. Consider the equation ; sin 5x . cos 3x = sin 6x . cos 2x.

(f) Solving equations by a change of variable :

(i) Equations of the form P sin cos ,sin .cos x x x x ± = b g 0 , where P(x, z) is a polynominal , can be solved by the change.cos x ± sin x = t ⇒ 1± 2sin x.cos x = t 2 . Consider the equation ; sin x + cos x = 1 + sin x . cos x .



(ii) Equation of the form of a .sin x + b . cos x + d = 0, where a ; b and d are real numberes and a b , ≠ 0 can be solved by changing sin x and cos x into their corresponding tangent of half the angle.Consider the equation 3cos x + 4 sin x = 5.

(iii) Many equations can be solved by introducing a new variable e.g. the equation

sin 4 2x + cos 4 2x = sin 2x. cos 2x changes to 2 1 1 2

0 ( ) y y + F H G I K J = by substituting,

sin 2x . cos 2x = y.

(g) Solving equations with the use of the Boundness of the functions sin x and cos x. Consider

the equation ; sin cos sin sin cos .cos x x x x x x 4

2 1 4

2 0 − F H G I K J + + − F H G I K J = .

5. SOME IMPORTANTPOINTS TO REMEMBER Ø While solving a trigonometric equation, squaring the equation at any step should be avoided

as far as possible. If squaring is necessary, check the solution for extraneous values. Ø Never cancel terms containing unknown terms on the two sides, which are in product. It

may cause loss of genuine solution.

Ø The answer should not contain such values of angles, which make any of the terms undefined.

Ø Domain should not be changed. If it is changed, necessary corrections must be incorporated.

Ø Check that the denominator is not zero at any stage while solving equations.

Ø Some times you may find that your answers differ from those in the package in their notations. This may be due to the different methods of solving the same problem. Whenever you come across such situation, you must check their authenticity. This will ensure that your answer is correct.

Ø While solving trigonometric equations you may get same set of solution repeated in your

answer. It is necessary for you to exclude these repetitions, e.g. nπ + 2 π , ) I n ( ∈ forms a

part of I k , 10 5 k

∈ π

+ π

the second part of the second set of solution (you can check by

putting k = 5 m + 2 (m∈I). Hence the final answer is I k , 10 5

k ∈

π +

π .

Ø Some times the two solution set consist partly of common values. In all such cases the common part must be presented only once. Now we present some illustrations for solving the different forms of trigonometric equations. Which will highlight the importance of above mentioned points.



DRILL EXERCISE 2

Solve the folloiwng trigonometric equations 1. cos sin cos sin θ θ θ θ + = + 2 2 2. sin 5x + sin 3x = sin 4x

3. sin 2 x + cos 4 x – 16 13

= 0 4. 4 sin 4 x + 12 cos 2 x = 7

5. ( tan )( sin ) tan 1 1 2 1 − + = + θ θ θ 6. sin x + sin 2x + sin 3 x = 1 2 2 cot x .

7. 2 cos 2x = 32 4 2

. cos x − . 8. sin 6 2x + cos 6 2x = 7/16.

9. sin 2 x + sin 2 2x = 1

Illustration 1: Solve: 7cos 2 θ + 3sin 2 θ = 4

Solution: Given 7cos 2 θ + 3sin 2 θ = 4 or, 7cos 2 θ + 3 (1 – cos 2 θ) = 4 or, 4cos 2 θ = 1

∴ cos 2 θ = 2

2 1

=

π 3

cos 2

⇒ θ = 3 n π ± π I n∈ ∀

Illustration 2: Solve: 3tan (θ 15 0 ) = tan (θ + 15 0 )

Solution: Given, 3tan (θ 15 0 ) = tan (θ + 15 0 )

or, ( ) ( ) 1

3 15 – tan 15 tan

0

0

= θ

+ θ

or, ( ) ( ) ( ) ( ) 2

4 15 tan 15 tan 15 tan 15 tan

0 0

0 0

= − θ − + θ

− θ + + θ (By componendo and dividendo)

or ( ) ( ) 2 15 15 sin

15 15 sin 0 0

0 0

= + θ − + θ

− θ + + θ

or, 2 sin2θ = 2 or sin2θ = 1 = sin 2 π

⇒ 2θ = nπ + (1) n 2 π

∴ θ = 2 nπ

+ ( 1) n 4 π , ∀ I n∈



Illustration 3:

Solve: cosθ cos2θ cos3θ = 4 1

Solution: 4cosθ cos2θ cos3θ = 1 or, (2cos3θ cosθ) 2cos2θ = 1 or, (cos4θ + cos2θ) 2cos2θ 1 = 0 or, 2cos4θ cos2θ + 2cos 2 2θ 1 = 0 or, 2cos4θ cos2θ + cos4θ = 0 or, cos4θ [2cos2θ + 1] = 0

If cos4θ = 0, 4θ = (2n + 1) 2 π

⇒ θ = (2n + 1) 8 π

If 2cos2θ + 1 = 0

or, cos2θ = – 2 1 = cos 3

2 m 2 2 3 2 π

± π = θ ⇒ π

∴ θ = mπ ± 3 π

Hence, θ = (2n + 1) 3 m or

8 π

± π π

where n, I m∈

Illustration 4: Solve: tanx + tan2x + tan3x = 0.

Solution: tanx + tan2x + tan3x = 0 or, tanx + tan2x + tan (x + 2x) = 0

or, tanx + tan2x + 0 x 2 tan x tan 1 x 2 tan x tan

= −

+

or, (tanx + tan2x) 0 x 2 tan x tan 1 1 1 =

−

+

If tanx + tan2x = 0, tanx = – tan2x or, tanx = tan(–2x) ⇒ x = nπ + (–2x) or, 3x = nπ

∴ x = 3 nπ

If 1 + x 2 tan x tan 1

1 −

= 0 then, 1 – tanx tan2x = – 1

or, tanx tan2x = 2 or, tanx ⋅ 2 x tan 1

x tan 2 2 = −

or, tan 2 x = 1 – tan 2 x

or, 2tan 2 x = 1 or, tan 2 x = 2 1

or, tanx = ± 2 1

or x = 2 1 tan m 1 − ± π



⇒ x = mπ + β

Hence x = , 2 1 tan m or

3 n 1

± π

π − I m , n ∈

Illustration 5 : Solve: 2sin 2 x – 5sinx cosx – 8cos 2 x = – 2.

Solution: In such problems we divide both sides by cos 2 x. This converts the given equation in a quadratic equation in tanx, which can be easily solved. Clearly, cosx ≠ 0 For if cosx = 0, then 2sin 2 x = – 2 ⇒ sin 2 x = – 1 which is impossible. Given equation is 2sin 2 x – 5sinx cosx – 8cos 2 x = – 2 or, 2tan 2 x – 5tanx – 8 = –2sec 2 x [dividing both sides by cos 2 x] or, 2tan 2 x – 5tanx – 8 + 2 (1 + tan 2 x) = 0 or, 4 tan 2 x –5tanx – 6 = 0 or, 4tan 2 x – 8tanx + 3tanx – 6 = 0 or, 4tanx (tanx –2) + 3 (tanx – 2) = 0 or, (tanx – 2) (4tanx + 3) = 0 ∴ either tanx – 2 = 0 ⇒ tanx = 2 = tanα (suppose) ⇒ x = nπ + α = nπ + tan –1 2

or, 4tanx + 3 = 0 ⇒ tanx = 4 3 − = tanβ (suppose)

⇒ x = nπ + β = mπ + tan –1

− 4 3 . where I m , n ∈

6. SOLVINGSIMULTANEOUS EQUATIONS Here we will discuss problems related to the solution of two equations satisfied simultaneously. We may divide the problems in two categories. (i) Two equations in one unknown (ii) Two equations in two unknowns.

Illustration 6: Find all values ofθ lying between 0 and 2π, satisfying the following equations, rsinθ = 3 and r + 4sinθ = 2 ( 3 + 1)

Solution: Given equations are,

rsinθ= 3 ... (i)

and r + 4sinθ = 2( 3 + 1) ... (ii) To find the value ofθ, we must eliminate r.

Now, from (i), r = θ sin 3



Substituting the value of r in (ii), we get,

) 1 3 ( 2 sin 4 sin 3

+ = θ + θ

or, 4sin 2 θ – 2 3 sinθ – 2sinθ + 3 = 0

or, 2sinθ (2sinθ – 3 ) – 1 (2sinθ – 3 ) = 0

or, (2sinθ – 3 ) (2sinθ – 1) = 0

If 2sinθ – 3 = 0, sin θ = 2 3 = sin

3 π

⇒ θ = nπ + (–1) n 3 π

If 2sinθ – 1 = 0, sinθ = 2 1 = sin ⇒

π 6

θ = nπ + (–1) n 6 π

Values of θ lying between 0 and 2π are 6 5 ,

3 2 ,

3 ,

6 π π π π

Illustration 7: Find the smallest positive values of x and y satisfying

x – y = ,4 π

and cotx + coty = 2 Solution:

Given x – y = 4 π

... (i) cotx + coty = 2 ... (ii)

From (ii), sin(x + y) = 2sinx.siny = cos (x – y) – cos(x + y)

= cos 4 π – cos (x + y)

⇒ sin (x + y) + cos (x + y) = cos 2 1

4 =

π

⇒ 2 1

sin (x + y) + 2 1

cos (x + y) = 2 1

⇒ cos (x + y – )4 π

= cos 3 π

⇒ x + y – 4 π = 2nπ ±

3 π

⇒ x + y = 2nπ ± 4 3 π

+ π

…. (iii)

for n = 0, x + y = ) 0 y , x ce (sin 12 7

> π

…. (iv)

From (i) and (iv), x = 6

y , 12 5 π

= π

Hence least positive values of x and y are 12 5π

and 6 π respectively. .



DRILL EXERCISE 3

Solve the following simultaneous equations

1. sin x sin y = 4 1 and cos x cos y = 4

3 2. y – x = 4

1 and cos (πx) cos (πy) =

2 2

3. cos (x – y) = 2 1 and cos (x + y) = – 2

1 4. sin x + cos y = 1 and x + y = 3

π

5. tan x + tan y = 2 and cos x cos y = 2 1

7. TRIGONOMETRIC INEQUATIONS To solve trigonometric inequation of the type f(x) ≤ a, or f(x) ≥ a where f(x) is some trigonometric ratio we take following steps. (i) Draw the graph of f(x) in a interval length equal to fundamental period of f(x). (ii) Draw the line y = a. (iii) Take the portion of the graph for which inequation is satisfied. (iv) To generalise add pn (n∈ I) and take union over set of integers , where p is fundamental

period of f(x).

Illustration 8: Find the solution set of the inequation sin x > 1/2.

Solution: When sinx = 1/2, the two values of x between 0 and 2π are π/6 and 5π/6.

π/6 5π/6

y

y = 1/2 x

From, the graph of y = sin x, it is obvous that, between 0 and 2π

sinx > 1/2 for 6 5 x

6 π

< < π

.

Hence sin x > 1/ 2 ⇒ 2nπ + π/6



Illustration 9:

Find the solution set of – 3 1



Illustration 10:

Solve: 2 / x 0 , x x x sin 2 x cos 2 2 2 2 2 π ≤ < + = −

Solution: In this problem, terms on the two sides of the equation are different in nature, L.H.S. is in trigonometric form, whereas R.H.S. is in algebraic form. Hence, we will use boundary conditions.

L.H.S. = 2 cos 2 x sin 2 x 2

= (1 + cosx) sin 2 x



Answer Key Drill Exercise 1

1. θ = mπ – 4 π , n ∈ I 2. θ = 2nπ + 6

5π , n ∈ I

3. θ = nπ + (–1) n

π 6 , n ∈ I 4.

0 6 3

2 3

5 6

, , , , & π π π π π

5. θ π π

= 3

5 3

, 6. θ = nπ ± 3 π , n ∈ I

Drill Exercise 2

1. θ π θ π π

= = + ∈ 2 2 3 6

n or m m n I , , 2. x = 4 nπ

, x = 2mπ ± 3 π , n, m∈ I

3. x = 2 nπ

± 6 π

4. x = 2nπ ± 4 π

5. n or m m n I π π π

− F H G I K J ∈ 4 , , 6. x

n n I = + ∈ 2 7 7

π π ,

7. x n n I = ∈ π, 8. x n n I = ± ∈ π π 4 12

,

9. (2n + 1) 6 π , I n∈

Drill Exercise 3

1. {(π (6n + 6k – 1)/6, π (6n – 6k – 1)/6), (π (6n + 6k + 1)/6, π(6n – 6k + 1)/6}, (n, k ∈ I).

2. {n, (4n + 1)/4, ((4n – 1)/4, n)} (n∈ I)

3. {(π (6k + 6n ± 1)/6, π (2k – 2n ± 1)/2) , (π (2k + 2n ± 1)/2, π(6k – 6n ± 1)/6}, (k, n ∈ I).

4. n 1 n 1 5 (( 1) sin 2 3 n, ( 1) sin 2 3 n 12 12

− − π π − − − + π − − − − π

, (n∈ I).

5. {(π (4m + 1)/4, π (8n – 4m + 1)/4)}, (n, m∈ I)



Drill Exercise 4

1.

π + π

π + π

∈ 6 11 n 2 ,

6 7 n 2

I n ∪ 2.

π + π π + π

∈ 2 n 2 , n 2

I n ∪

3. (4πn, π (12n + 1)/3 ∪ (π (12n + 1)/3, 2π + 4πn), (n ∈ I)

4. n I

5 2n , 2n 2n , (2n 1) 6 6 ∈

π π π π + ∪ π + + π ∪ 5. R –

∈

π I n , 4 n

Drill Exercise 5

1. I n , n 2 x ∈ π + π = , I n ,2

n x ∈ π + π = 2. x = 0

3. φ 4. x = (2n + 1) 2 π , y = mπ, z = (2t + 1) 2

π , (n, m, t ∈ I)

5. x = nπ, y = (4m + 1) 3 π , (n, m∈ I)

Trigonometric Equation 1€¦ · E The general solution should be given unless the solution is...

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