CHAPTER 14 Solutions of Trignometric Equation

[Pages:5]version: 1.1

CHAPTER

14 Solutions of Trignometric

Equation

114. .QSuoaludtriaontiscoEfqTuriagtoiononms etric Equations

eLearn.Punjab

eLearn.Punjab

14.1 Introduction

The Equations, containing at least one trigonometric function, are called Trigonometric Equations, e.g., each of the following is a trigonometric equation:

= sin x 2 ,= Sec x ta-n x = an+d sin2 x sec x 1 3

5

4

Trigonometric equations have an ininite number of solutions due to the periodicity of the trigonometric functions. For example

If sinq =q then q = 0, ? , ? 2 ,... which can be wr= itten as q n , where n Z.

In solving trigonometric equations, irst ind the solution over the interval whose length is equal to its period and then ind the general solution as explained in the following examples:

Example 1: Solve the equation sin x = 1 2

Solution: sin x = 1 2

a sin x is positive in I and II Quadrants with the reference angle x = . 6

= x and = x= - 5 ,

6

66

where x [0, 2 ]

General values of x are + 2n and 5 + 2n , n Z

6

6

Hence solution set =6 + 2n 56 + 2n

,nZ

version: 1.1

2

114. .QSuoaludtrioantiscoEfqTuriagtoinoonms etric Equations

Example 2: Solve the equation: 1 + cos x = 0

Solution: 1 + cos x = 0

cos x = -1

Since cos x is -ve, there is only one solution x = p

Since 2p is the period of cos x

General value of x is p + 2np,

n Z

Hence solution set = {p + 2np},

nZ

in [0, 2p]

Example 3: Solve the equation: 4 cos2x - 3 = 0

Solution: 4 cos2 x - 3 = 0

cos2 x = 3 ? cos x = 3

4

2

i.

If cos x = 3

2

Since cos x is +ve in I and IV Quadrants with the reference angle

x= 6

= = x -and x = 2 6

11 66

As 2p is the period of cos x.

where x [0, 2 ]

General value of x are + 2n and 11 + 2n , n Z

6

6

ii. if cos x = - 3 2

Since cos x is -ve in II and III Quadrants with reference angle x = 6

x = - =5 and x = x + = 7

66

66

where x [0, 2 ]

As 2p is the period of cos x.

3

eLearn.Punjab

eLearn.Punjab version: 1.1

114. .QSuoaludtriaontiscoEfqTuriagtoiononms etric Equations

eLearn.Punjab

eLearn.Punjab

General values of x are 5 + 2n and 7 + 2n , n Z

6

6

Hence solution set =6 + 2n 116 + 2n 56 + 2n 76 + 2n

14.2 Solution of General Trigonometric Equations

When a trigonometric equation contains more than one trigonometric functions, trigonometric identities and algebraic formulae are used to transform such trigonometric equation to an equivalent equation that contains only one trigonometric function.

The method is illustrated in the following solved examples:

Example 1: Solve: sin x + cos x = 0.

Solution: sin x + cos x = 0

sin x + cos x cos x cos x

= 0 (Dividing by cos x 0)

tan x +1 = 0

t-an x = 1

a tan x is -ve in II and IV Quadrants with the reference angle

x= 4

x = - =3 , 44

where x [0, ]

As p is the period of tan x,

General value of x is 3 + n , 4

nU Z

Solution set =

3 4

+

n

,n U Z.

version: 1.1

4

114. .QSuoaludtrioantiscoEfqTuriagtoinoonms etric Equations

eLearn.Punjab

eLearn.Punjab

Example 2: Find the solution set of: sin x cos x = 3 . 4

Solution: sin x cos x = 3 . 4

1 2

(

2sin

x

cos

x

)

= 3 4

sin 2x = 3 2

a sin 2x is +ve in I and II Quadrants with the reference angle 2x = 3

2x = and 2x = 3

- =2 33

are two solutions in [0,2

]

As 2p is the period of sin 2x .

General values of 2x are + 2n and 2 + 2n , , n U Z

3

3

General values of x are + n and + n

,

6

3

Hence solution set = = 6 + n 3 + n

nU Z , nU Z

Note: In solving the equations of the form sin kx = c, we irst ind the solution pf sin u = c (where kx = w) and then required solution is obtained by dividing each term of this solution set by k.

Example 3: Solve the equation: sin 2x = cos 2x

Solution:

sin2x = cos2x

2sinx cos x = cosx

2sinx cos x - cosx = 0

cosx(2sinx - 1) = 0

version: 1.1

5

114. .QSuoaludtriaontiscoEfqTuriagtoiononms etric Equations

cosx = 0 or 2sinx - 1 = 0 i. If cosx = 0

x= 2

and x = 3 2

where x U [0,2p]

As 2p is the period of cos x .

General values of x are p + 2np and 3p + 2np, nUZ,

ii. If 2 sin x - 1 = 0

2

2

sin x = 1

2

Since sin x is +ve in I and II Quadrants with the reference angle x = p

6

x = p and x =p - p = 5p where x U [0, 2p]

6

66

As 2p is the period of sin x.

General values of x are and p + 2np and 5 p + 2np, nUZ,

6

6

Hence

solution

set

=

p2

+

2np

3p 2

+ 2np p6

+

2np

5

p 6

+ 2np ,

nz

Example 4: Solve the equation: sin2 x + cos x = 1.

Solution:

sin2 x + cos x = 1

1 - cos2 x + cos x = 1

- cos x (cos x - 1) = 0

cos x = 0 or cos x - 1 = 0

i. If cos x = 0

x = p and x = 3p ,

2

2

where x U [0, 2p]

6

eLearn.Punjab

eLearn.Punjab version: 1.1

114. .QSuoaludtrioantiscoEfqTuriagtoinoonms etric Equations

eLearn.Punjab

eLearn.Punjab

As 2p is the period of cos x

General values of x are p + 2np and 3p + 2np , nUZ

2

2

ii. If cos x = 1

x = 0 and x = 2p ,

where x U [0, 2p]

As 2p is the period of cos x

General values of x are 0 + 2np and 2p + 2np, nUZ.

Solution Set =p2 + 2np 32p + 2np {2np} {2p + 2np},n z {2(n +1)p} {2np}, n z

Hence

the

solution

set

=

p2

+

2np

3p 2

+ 2np {2np},n z

Sometimes it is necessary to square both sides of a trigonometric equation. In such

a case, extaneous roots can occur which are to be discarded. So each value of x must be

checked by substituting it in the given equation. For example, x = 2 is an equation having a root 2. On squaring we get x2 - 4 which gives

two roots 2 and -2. But the root -2 does not satisfy the equation x = 2. Therefore, -2 is an extaneous root.

Example 5: Solve the equation: csc=x 3 + cot x.

Solution: csc=x 3 + cot x

1 = 3 + cos x

sin x

sin x

=1 3 sin x + cos x

.......(i)

1 - cos x =3 sin x

(1 - cos x)2 = ( 3 sin x)2

version: 1.1

7

114. .QSuoaludtriaontiscoEfqTuriagtoiononms etric Equations

eLearn.Punjab

eLearn.Punjab

1 - 2cos x + cos2 x =3sin2 x

1 - 2cos x + cos2 x = 3(1 - cos2 x)

4cos2 x - 2cos x - 2 =0

2cos2 x - cos x -1 =0

(2cos x +1)(cos x -1) =0 co-s x = 1 or = cos x 1

2

i.

If cos x = - 1

2

Since cos x is -v e in II and III Quadrants with the reference angle x = p 3

x =p - p = 2p 33

and x =p + p = 4p , where x U [0, 2p]

33

Now x = 4p does not satisfy the given equation (i). 3

x = 4p is not admissible and so x = 2p is the only solution.

3

3

Since 2p is the period of cos x

General value of x is 2p + 2np , nUZ

ii. If cos x = 1

3

x = 0 and x = 2p where x U [0, 2p]

Now both csc x and cot x are not defined for x = 0 and x = 2

x = 0 and x = 2 are not admissible.

Hence solution set =

2p 3

+

2np

, nUZ

8

version: 1.1

114. .QSuoaludtrioantiscoEfqTuriagtoinoonms etric Equations

eLearn.Punjab

eLearn.Punjab

Exercise 14

1. Find the solutions of the following equations which lie in [0, 2p]

i) sin x = - 3

ii) cosecq = 2

iii) sec x = -2

iv)

2

2. Solve the following trigonometric equations:

i)

tan2 q = 1

ii) cos ec2q = 4

iii) sec2 q = 4

iv)

3

3

3

Find the values of q satisfying the following equations:

3. 3tan2 q + 2 3 tanq +1 =0 4. tan2 q - secq -1 =0 5. 2sinq + cos2 q -1 =0 6. 2sin2 q - sinq = 0

7. 3cos2 q - 2 3 sinq cosq - 3sin2 q = 0 [Hint: Divide by sin2q]

Find the solution sets of the following equations:

8. 4 sin2q - 8cosq + 1 = 0

9.

3 tan x - sec x -1 =0

10. cos 2x = sin 3x

11. sec 3q= secq

12. tan 2q + cotq = 0

[Hint: sin3x = 3sinx - 4sin3x]

13. sin 2x + sinx = 0

14. sin 4x - sin 2x = cos 3x

15. sin x + cos 3x = cos 5x

16. sin 3 x + sin 2x + sin x = 0

17. sin 7x - sin x = sin 3 x

18. sin x + sin 3x + sin 5x = 0

19. sin q + sin 3q + sin 5q + sin 7q = 0 20. cos q + cos 3q + cos 5 q + cos 7q = 0

cotq = 1 3

cot2 q = 1 3

version: 1.1

9

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download