Subject : Mathematics Topic : TRIGONOMETRI EQUATIONS - TEKO CLASSES

Page : 1 of 15 TRIG. EQUATIONS

fo/u fopkjr Hkh# tu] ugha vkjEHks dke] foifr n[s k NkMs +s rjq ra e/;e eu dj ';keA i#q "k flga lda Yi dj] lgrs foifr vuds ] ^cuk^ u NkMs +s /;;s dk]s j?kcq j jk[ks Vds AA

jfpr% ekuo /keZ iz.ksrk ln~xq# Jh j.kNksM+nklth egkjkt

STUDY PACKAGE

Subject : Mathematics Topic : TRIGONOMETRI EQUATIONS

R

TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779, 98930 58881 , BHOPAL

FREE Download Study Package from website:

Index 1. Theory 2. Short Revision 3. Exercise (Ex. 1 + 5 = 6) 4. Assertion & Reason 5. Que. from Compt. Exams 6. 39 Yrs. Que. from IIT-JEE(Advanced) 7. 15 Yrs. Que. from AIEEE (JEE Main)

Student's Name :______________________

Class

:______________________

Roll No.

:______________________

Address : Plot No. 27, III- Floor, Near Patidar Studio,

Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal

: 0 903 903 777 9, 98930 58881, WhatsApp 9009 260 559





Page : 2 of 15 TRIG. EQUATIONS

Trigonometric Equation

1 . Trigonometric Equation :

An equation involving one or more trigonometric ratios of an unknown angle is called a trigonometric

equation.

2 . Solution of Trigonometric Equation :

A solution of trigonometric equation is the value of the unknown angle that satisfies the equation.

e.g. if sin = 1 2

3 9 11 = 4 , 4 , 4 , 4 , ...........

Thus, the trigonometric equation may have infinite number of solutions (because of their periodic nature) and

can be classified as :

(i)

Principal solution

(ii) General solution.

2.1 Principal solutions:

The solutions of a trigonometric equation which lie in the interval

[0, 2) are called Principal solutions.

1 e.g Find the Principal solutions of the equation sinx = 2 . Solution.

TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779, 98930 58881 , BHOPAL

FREE Download Study Package from website:

1

sinx =

2

there exists two values

5

1

i.e.

6 and 6 which lie in [0, 2) and whose sine is 2

1

Principal solutions of the equation sinx = 2 are 6 ,

2.2 General Solution :

5 6 Ans.

The expression involving an integer 'n' which gives all solutions of a trigonometric equation is called General solution.

General solution of some standard trigonometric equations are given below.

3 . General Solution of Some Standard Trigonometric Equations :

(i) If sin = sin

(ii) If cos = cos

(iii) If tan = tan

(iv) If sin? = sin?

(v) If cos? = cos?

(vi) If tan? = tan?

Some Important deductions :

(i) sin = 0

(ii) sin = 1

(iii) sin = ? 1

(iv) cos = 0

(v) cos = 1

(vi) cos = ? 1

(vii) tan = 0

= n + (-1)n = 2n ?

= n+ = n ? , n . = n ? , n . = n ? , n .

where

-

2

,

2

,

where [0, ],

where - , , 2 2

n . n . n .

[ Note: is called the principal angle ]

= n,

n

= (4n + 1) 2 , n

= (4n ? 1) , n

2

= (2n + 1) , n

2

= 2n,

n

= (2n + 1), n

= n,

n

Solved Example # 1 Solve sin = 3 . 2

Solution.

Page : 3 of 15 TRIG. EQUATIONS

sin = 3

2

sin = sin 3

= n + (? 1)n 3 , n

Solved Example # 2

2 Solve sec 2 = ? 3 Solution.

2

sec 2 = ?

3

cos2 = ? 3

2

2 = 2n ? 5 , n

6

5 = n ? 12 , n

Ans.

5 cos2 = cos 6 Ans.

TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779, 98930 58881 , BHOPAL

FREE Download Study Package from website:

Solved Example # 3

Solve tan = 2

Solution.

Let

tan = 2

............(i)

2 = tan

tan = tan

= n + , where = tan?1(2), n

Self Practice Problems:

1.

Solve cot = ? 1

1

2.

Solve cos3 = ? 2

2n 2

Ans. (1)

= n ? 4 , n

(2)

3 ? 9 ,n

Solved Example # 4

1 Solve cos2 = 2 Solution.

1

cos2 =

2

cos2 =

1 2

2

cos2 = cos2

4

= n ? 4 , n Ans.

Solved Example # 5

Solve 4 tan2 = 3sec2

Solution.

4 tan2 = 3sec2

.............(i)

For equation (i) to be defined (2n + 1) 2 , n

equation (i) can be written as:

4 sin2

3

cos2 = cos2

4 sin2 = 3

(2n + 1) 2 , n

cos2 0

sin2 =

3 2 2

Page : 4 of 15 TRIG. EQUATIONS

sin2 = sin2 3

= n ? 3 , n Ans.

Self Practice Problems :

1.

Solve 7cos2 + 3 sin2 = 4.

2.

Solve 2 sin2x + sin22x = 2

Ans. (1)

n ? 3 , n

(2) (2n + 1) , n

or

n ? , n

2

4

Types of Trigonometric Equations :

TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779, 98930 58881 , BHOPAL

FREE Download Study Package from website:

Type -1

Trigonometric equations which can be solved by use of factorization.

Solved Example # 6

Solve (2sinx ? cosx) (1 + cosx) = sin2x.

Solution.

(2sinx ? cosx) (1 + cosx) = sin2x

(2sinx ? cosx) (1 + cosx) ? sin2x = 0

(2sinx ? cosx) (1 + cosx) ? (1 ? cosx) (1 + cosx) = 0

(1 + cosx) (2sinx ? 1) = 0

1 + cosx = 0

or

2sinx ? 1 = 0

cosx = ? 1

1

or

sinx =

2

x = (2n + 1), n or

sin x = sin 6

Solution of given equation is

(2n + 1), n Self Practice Problems :

or

n + (?1)n , n

6

x = n + (? 1)n

6 , n

Ans.

x

1.

Solve cos3x + cos2x ? 4cos2 = 0

2

2.

Solve cot2 + 3cosec + 3 = 0

Ans. (1) (2)

Type - 2

(2n + 1), n 2n ? , n or

2

n + (?1)n + 1 , n 6

Trigonometric equations which can be solved by reducing them in quadratic equations.

Solved Example # 7

Solve 2 cos2x + 4cosx = 3sin2x

Solution.

2cos2x + 4cosx ? 3sin2x = 0 2cos2x + 4cosx ? 3(1? cos2x) = 0 5cos2x + 4cosx ? 3 = 0

cos

x

-

-

2

+ 5

19

cos

x

-

-

2

- 5

19

= 0

cosx [? 1, 1] x R

........(ii)

- 2 - 19

cosx

5

equation (ii) will be true if

- 2 + 19

cosx =

5

- 2 + 19

cosx = cos, where cos =

5

x = 2n ? where

= cos?1

-

2

+ 5

19 , n

Ans.

Page : 5 of 15 TRIG. EQUATIONS

Self Practice Problems : 1.

Solve

cos2 ? (

2 + 1) cos -

1 2

= 0

2.

Solve

Ans.

Type - 3

4cos ? 3sec = tan

(1)

2n ? , n

3

or

2n ? , n

4

(2)

n + (? 1)n

where

=

sin?1

- 1- 8

17

,

n

or

n + (?1)n

where

=

sin?1

- 1+ 8

17

,

n

Trigonometric equations which can be solved by transforming a sum or difference of trigonometric ratios into their product.

Solved Example # 8 Solve cos3x + sin2x ? sin4x = 0

Solution.

cos3x + sin2x ? sin4x = 0 cos3x ? 2cos3x.sinx = 0

cos3x = 0

3x = (2n + 1) 2 , n

x = (2n + 1) 6 , n solution of given equation is

(2n + 1) 6 , n

or

cos3x + 2cos3x.sin(? x) = 0

cos3x (1 ? 2sinx) = 0

or

1 ? 2sinx = 0

1

or

sinx = 2

or

x = n + (?1)n 6 , n

n + (?1)n 6 , n

Ans.

Self Practice Problems :

1.

Solve sin7 = sin3 + sin

TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779, 98930 58881 , BHOPAL

FREE Download Study Package from website:

2.

Solve 5sinx + 6sin2x +5sin3x + sin4x = 0

3.

Solve cos ? sin3 = cos2

Ans. (1) (2) (3)

n

3 ,n

or

n

2 ,n

or

2n 3 , n or

n 2 ? 12 , n

2 2n ? 3 , n

2n ? 2 , n

or

n + 4 , n

Type - 4 Trigonometric equations which can be solved by transforming a product of trigonometric ratios into their sum or difference.

Solved Example # 9

Solve sin5x.cos3x = sin6x.cos2x

Solution.

sin5x.cos3x = sin6x.cos2x

sin8x + sin2x = sin8x + sin4x

2sin2x.cos2x ? sin2x = 0

sin2x = 0

or

2cos2x ? 1 = 0

2x = n, n or

1 cos2x = 2

x = n , n or 2

2x = 2n ? 3 , n

x

=

n

?

6,

n

Solution of given equation is

Type - 5

n , n 2

or

n ?

6

, n

2sin5x.cos3x = 2sin6x.cos2x sin4x ? sin2x = 0 sin2x (2cos2x ? 1) = 0

Ans.

Trigonometric Equations of the form a sinx + b cosx = c, where a, b, c R, can be solved by dividing both sides of the equation by a2 + b2 .

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

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

Google Online Preview   Download