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STUDY PACKAGE

Subject : Mathematics Topic: Trigonometric Equation & Properties & Solution of Triangle

Index

?

1. Theory

2. Short Revision

3. Exercise

4. Assertion & Reason

5. Que. from Compt. Exams

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Page : 2 of 29

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

=

,

4

3 ,

4

9 ,

4

11 , ...........

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 GROUP MATHS BY SUHAAG SIR PH: (0755)- 32 00 000, 98930 58881

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)

2

,n

= (2n + 1)

2

,

n

= 2n,

n

= (2n + 1), n

= n,

n

Solved Example # 1

Solve

sin =

3 .

2

Solution.

Page : 3 of 29

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.

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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

Ans. (1)

= n ?

4

,

n

(2)

2n 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 29

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)

2

,

n

or

n ?

4

,

n

Types of Trigonometric Equations :

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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 6 , n

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

cosx =

-2+ 5

19

cosx = cos,

where

cos =

-2+ 5

19

x = 2n ? where

= cos?1

-

2

+ 5

19 , n

Ans.

Page : 5 of 29

Self Practice Problems : 1.

Solve

cos2 ? (

2 + 1) cos -

1 2

= 0

2.

Solve 4cos ? 3sec = tan

Ans. (1)

2n ? , n 3

or

2n ? , n

4

(2)

n + (? 1)n

where

=

sin?1

-

1- 8

17

,

n

Type - 3

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.

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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

Self Practice Problems :

1.

Solve sin7 = sin3 + sin

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.

2.

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

3.

Solve cos ? sin3 = cos2

n

Ans. (1)

3 ,n

or

n 2

?

12

,

n

(2)

n 2

,

n

or

2 2n ? 3 , n

2n

(3)

3 , n or

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

1

2x = n, n or

cos2x = 2

x

=

n 2

,

n

or

2x = 2n ? 3 , n

Type - 5

x = n ? 6 , n Solution of given equation is

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 .

Page : 6 of 29

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Solved Example # 10

Solve sinx + cosx = 2

Solution.

sinx + cosx = 2

Here a = 1, b = 1.

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

divide both sides of equation (i) by 2 , we get

1

1

sinx .

+ cosx. = 1

2

2

sinx.sin + cosx.cos = 1

4

4

cos

x -

4

=

1

x ? 4 = 2n, n

x = 2n +

4

,

n

Solution of given equation is

2n +

4

,

n

Ans.

Note : Trigonometric equation of the form a sinx + b cosx = c can also be solved by changing sinx and cosx into their corresponding tangent of half the angle.

Solved Example # 11

Solve 3cosx + 4sinx = 5

Solution.

Let

3cosx + 4sinx = 5

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

1- tan2 x

cosx =

2 1+ tan2 x

&

2

equation (i) becomes

2 tan x

sinx

=

2 1+ tan2 x

2

3

1- tan2 1+ tan2

x

2

x 2

+

4

2 tan x

2

1+

tan2

x 2

=

5

x tan 2 = t equation (ii) becomes

........(ii)

3

1 - 1+

t2 t2

+ 4

2 t 1+ t2

= 5

4t2 ? 4t + 1 = 0

(2t ? 1)2 = 0

1 t =

2

x t = tan

2

x 1 tan 2 = 2

x

1

tan 2 = tan, where tan = 2

x 2 = n +

x = 2n + 2

where

=

tan?1

1 2

,

n

Ans.

Self Practice Problems :

1.

Solve 3 cosx + sinx = 2

Page : 7 of 29

x

2.

Solve sinx + tan 2 = 0

Ans. (1)

2n + 6 , n

(2) x = 2n, n

Type - 6

Trigonometric equations of the form P(sinx ? cosx, sinx cosx) = 0, where p(y, z) is a polynomial, can be solved by using the substitution sinx ? cosx = t.

Solved Example # 12

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Solve sinx + cosx = 1 + sinx.cosx

Solution.

Let

sinx + cosx = 1 + sinx.cosx

sinx + cosx = t sin2x + cos2x + 2 sinx.cosx = t2

........(i)

sinx.cosx = t2 -1

2

Now put

sinx + cosx = t and sinx.cosx = t2 -1 in (i), we get 2

t = 1 + t2 -1

2

t2 ? 2t + 1 = 0

t = 1

sinx + cosx = 1

t = sinx + cosx

.........(ii)

divide both sides of equation (ii) by 2 , we get

1

1

1

sinx. + cosx. =

2

2

2

cos x - 4

=

cos

4

x ?

4

= 2n ?

4

(i)

if we take positive sign, we get

x = 2n + 2 , n Ans.

(ii) if we take negative sign, we get

x = 2n, n

Ans.

Self Practice Problems:

1.

Solve sin2x + 5sinx + 1 + 5cosx = 0

2.

Solve 3cosx + 3sinx + sin3x ? cos3x = 0

3.

Solve

Ans.

Type - 7

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

(1)

n ? 4 , n

(2)

n ? 4 , n

(3)

2n +

2

,

n

or

2n, n

or

n +

4

,

n

Trigonometric equations which can be solved by the use of boundness of the trigonometric ratios sinx and cosx. Solved Example # 13

Solve

sinx cos

x 4

- 2 sin x

+

1 +

sin

x 4

- 2cos x cos x

= 0

Solution.

sinx cos

x - 2 sin x

4

+

1+ sin x - 2cos x cos x

4

= 0

.......(i)

x

x

sinx.cos ? 2sin2x + cosx + sin .cosx ? 2cos2x = 0

4

4

sin x.cos

x 4

+ sin x .cos x

4

?

2

(sin2x

+

cos2x)

+

cosx

=

0

sin 5x + cosx = 2

4

........(ii)

Page : 8 of 29

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Now equation (ii) will be true if

5 x sin = 1

4

and cosx = 1

5 x 4

= 2n +

2

,

n

and

x = 2m, m

x

=

(8n

+ 5

2)

,

n

........(iii)

and

x = 2m, m

Now to find general solution of equation (i)

(8n + 2) 5

= 2m

8n + 2 = 10m

5m - 1

n= 4

if

m = 1

if

m = 5

.........

.........

if

m = 4p ? 3, p

then then ......... ......... then

n = 1 n = 6 .........

......... n = 5p ? 4, p

........(iv)

general solution of given equation can be obtained by substituting either m = 4p ? 3 in

equation (iv) or n = 5p ? 4 in equation (iii)

general solution of equation (i) is

(8p ? 6), p

Ans.

Self Practice Problems :

1.

Solve sin3x + cos2x = ? 2

2.

Solve 3 sin 5x - cos2 x - 3 = 1 ? sinx

Ans. (1)

(4p ? 3)

2

,

p

(2)

2m +

2

,

m

Exercise -1

Part : (A) Only one correct option

1.

The solution set of the equation 4sin.cos ? 2cos ? 2 3 sin + 3 = 0 in the interval (0, 2) is

(A)

3 4

,

7 4

(B)

3

,

5 3

(C)

3 4

,

,

, 3

5 3

(D)

6

,

5 , 6

11 6

2.

All solutions of the equation, 2 sin + tan = 0 are obtained by taking all integral values of m and n in:

2 (A) 2n + 3 , n

2 (B) n or 2m ? 3 where n, m

(C) n or m ? 3 where n, m

(D) n or 2m ? 3 where n, m

7

3.

If 20 sin2 + 21 cos - 24 = 0 & 4 < < 2 then the values of cot 2 is:

(A) 3

15 (B) 3

15 (C) - 3

4.

The general solution of sinx + sin5x = sin2x + sin4x is:

(A) 2 n ; n

(B) n ; n

(C) n/3 ; n

(D) - 3 (D) 2 n/3 ; n

1

5.

A triangle ABC is such that sin(2A + B) = . If A, B, C are in A.P. then the angle A, B, C are

2

respectively.

(A)

5 12

,

4

,

3

(B)

4

,

3,

5 12

(C)

3,

4

,

5 12

(D)

3,

5 12

,

4

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