Trigonometric Equations : Session 1 Illustrative Problem Solve :sinx + cosx = 2 Solution:

Trigonometric Equations : Session 1

Illustrative Problem

Solve :sinx + cosx = 2

Solution:

no solution.

A trigonometric equation

is an equation

Contains trigonometric functions of variable angle

sin = ½

2 sin2 + sin22 = 2.

Definition

For sin = ½ , = /6, 5 /6, 13 /6,…….

Solution of Trigonometric Equation:

Values of , which satisfy the trigonometric equation

No. of solutions are infinite .

Why ? - Periodicity of trigonometric functions.

e.g. - sin, cos have a period as 2

Periodicity and general solution

Periodicity of trigonometric functions.

sin have a period 2

f(+T) = f()

sin

0 2 3 4

Periodicity and general solution

Graph of y=sinx

sinx is periodic of period 2

0

32

-2 232

2

2

-

Y

X

(0,1)

(0,-1)

Graph of y=cosx

(0,1)

-2 232

2

2

32

X

0

(0,-1)

Y

cosx is periodic of period 2

tanx is periodic of period

tanx is not defined at x=(2 where n is integer

n+1)2

Graph of y=tanx

-2 232

2

2

32

X

0

Y

For sin = ½ , = /6, 5/6, 13/6,…….

As solutions are infinite , the entire set of solution can be written in a compact form.

This compact form is referred to as general solution

Or = n +(-1)n ( /6)General Solution

Periodicity and general solution

Principal Solutions

Solutions in 0x2

principal solutions.

Illustrative problem

Find the principal solutions of

tanx = 1

3

Solution

We know that tan( - /6) = 1

3

and tan(2 - /6) = 1

3

principal solutions are 5/6 and 11/6

Illustrative problem

Find the principal solution of the equation sinx = 1/2

Solution

sin/6 = 1/2

and sin(- /6) = 1/2

principal solution are x = /6 and 5/6.

sin = PM/OP

For sin = 0 , PM = 0

For PM = 0, OP will lie on XOX’Y

X

O

P

M

X’

Y’ is an integral multiple of π.

= 0, ±π, ±2π, ±3π …..

General solution of sin = 0

Or = nπ , n є Z (n belongs to set of integers)

For sin = 0 ,

is an integer multiple of π.

Hence, general solution of sin = 0 is

= nπ , where n є Z,

General solution of sin = 0

cos = OM/OP

For cos = 0 , OM = 0

For OM = 0, OP will lie on YOY’

is an odd integer multiple of π/2.

= ±π/2, ±3π/2, ±5π/2….

Y

X

O

P

M

X’

Y’

General solution of cos = 0

Or = (2n+1)π/2 , n є Z (n belongs to set of integers)

For cos = 0 ,

is an odd integer multiple of π/2.

Hence, general solution of cos = 0 is

= (2n+1)π/2 , where n є Z,

General solution of cos = 0

tan = PM/OM

For tan = 0 , PM = 0

For PM = 0, OP will lie on XOX’ Y

X

O

P

M

X’

Y’

is an integer multiple of π.

= 0,±π, ±2π, ±3π….

Same as sin = 0

General solution of tan = 0

Or = nπ , n є Z (n belongs to set of integers)

For tan = 0 ,

is an integer multiple of π.

Hence, general solution of tan = 0 is

= nπ, where n є Z,

General solution of tan = 0

Find the general value of x

satisfying the equation sin5x = 0

Solution:

sin5x = 0 = sin0

=> 5x = n

=> x = n/5

=>x = n/5 where n is an integer

Illustrative Problem

If sin = k -1 k 1

Let k = sin, choose value of between –/2 to /2

If sin = sin sin - sin = 0

02

sin2

cos2

General solution of sin = k

02

sin2

cos2

02

cos

02

sin

2)1n2(

2

= (2n+1)π -

n2

= 2nπ +

= nπ +(-1)n , where n є Z

Odd , -ve Even , +ve

General solution of sin = k

If cos = k -1 k 1

Let k = cos, choose value of between 0 to

If cos = cos cos - cos = 0

02

sin2

sin2

General solution of cos = k

02

sin

02

sin

2

n2

= 2nπ -

n2

= 2nπ +

= 2nπ , where n є Z

-ve +ve

02

sin2

sin2

General solution of cos = k

If tan = k - < k <

Let k = tan, choose value of between - /2 to /2

If tan = tan tan - tan = 0

sin.cos - cos.sin = 0

General solution of tan = k

2

sin( - ) = 0

- = nπ , where n є Z

= nπ +

sin.cos - cos.sin = 0

= nπ+ , where n є Z

General solution of tan = k

Illustrative problem

Find the solution of sinx = 32

Solution:

nx n ( 1)3

As sinx sin3

Illustrative problem

Solution:

Solve tan2x = cot(x )6

We have tan2x =

cot(x )6

tan( x )2 6

2tan( x)

3

22x n x

3

2x n ,where n isan int eger

3

Illustrative problem

Solution:

Solve sin2x + sin4x + sin6x = 0

2sin4x cos2x sin4x 0

sin4x(2cos2x 1) 0

1sin4x 0or cos2x

2

2sin4x 0or cos2x cos

3

nx or x n ,where n is an int eger

4 3

24x n or 2x 2n ,where n isan integer

3

Illustrative Problem

Solution:

Solve 2cos2x + 3sinx = 0

22cos x 3sinx 0

22(1 sin x) 3sinx 0

(2sinx 1)(sinx 2) 0 1

sinx or sinx 22

1 7sinx sin

2 6

n 7x n ( 1) where n is an int eger

6

General solution of sin2x = sin2 cos2x = cos2, tan2x = tan2

n where n is an integer.

Illustrative Problem

Solve : 4cos3x-cosx = 0

Solution:34cos x cosx 0

2cosx(4cos x 1) 0

2cosx 0 or 4cos x 1

22 21

x (2n 1) or cos x cos2 2 3

x (2n 1) or n / 3, n Z2

Illustrative Problem

Solve :sinx + siny = 2

Solution:

sinx siny 2

sinx 1 and siny 1

sinx sin and siny sin2 2

n mx n ( 1) and y m ( 1)2 2

where n and m are int eger.

Class Exercise Q1.

Solve :sin5x = cos2x

Solution:

cos2x sin5x

cos2x cos( 5x)2

2x 2n ( 5x)2

taking positive sign

7x 2n2

(4n 1)x

14

taking negative sign

2x 2n 5x2

(4n 1)x , n I

6

Class Exercise Q2.

Solve :2sinx + 3cosx=5

Solution:

2sinx 3cosx 5 is possible only when

sinx and cosx attains their maximum

value i.e.sinx 1 and cosx 1.

both sinx and cosx cannot be 1 for any value of x.

hence no solution.

Class Exercise Q3.

Solve :7cos2 +3sin2 = 4

Solution:

2 27cos 3(1 cos ) 4

2 27cos 3 3cos 4

2 1cos

4 2 2cos cos

3

n , n I3

Class Exercise Q4.

Solution:

Solve : 2cos2 2sin 2

2cos2 2sin 2

2sin 2(cos2 1) 0 22sin 4sin 0

322sin (1 2 2 sin ) 0

1sin 0 and sin

2

nn , n ( 1) where n I

6

Class Exercise Q5.

Show that 2cos2(x/2)sin2x = x2+x-2 for 0<x</2 has no real solution.

Solution:

2 2xL.H.S. 2cos sin x

2

2(1 cosx).sin x 2

22

1R.H.S x

x

21

x 2 2x

Hence no solution

Class Exercise Q6.

Solution:

Find the value(s) of x in (- , ) which satisfy the following equation

2 31 cos x cos x cos x ....to 38 4

2 31 cos x cos x cos x ....to 28 8

2 31 cosx cos x cos x ....to 2

12

1 cosx

1

cosx2

Class Exercise Q6.

Solution:

Find the value(s) of x in (- , ) which satisfy the following equation

2 31 cos x cos x cos x ....to 38 4

1COSX

2

2x , x

3 3

Class Exercise Q7.

Solve the equation

sinx + cosx = 1+sinxcosx

Solution:

Let sinx cosx z

squaring both sides

21 2sinx.cosx z 2(z 1)

sinx.cosx2

2z 1

sinx cosx 12

2z 2z 1 0 2(z 1) 0 z 1

Class Exercise Q7.

Solve the equation

sinx + cosx = 1+sinxcosx

Solution:

sinx cosx 1

1 1 1cosx sinx

2 2 2 1

cos(x ) cos4 42

x 2n4 4

(4n )x 2n , , n I

2

2(z 1) 0 z 1

Class Exercise Q8.

Solution:

3r 4(1 sinx)

sinx 2

2

3 4sinx 4sin x

4sin x 4sinx 3 0

If rsinx=3,r=4(1+sinx),

then x is(0 x 2 )

(a) or (b) or3 4 2

7 5(c) or (d) or

6 6 6 6

(2sinx 1)(2sinx 3) 0

Class Exercise Q8.If rsinx=3,r=4(1+sinx),

then x is

Solution:

(a) or (b) or3 4 2

(0 x 2 )

7 5(c) or (d) or

6 6 6 6

(2sinx 1)(2sinx 3) 0

5x or

6 6

1 3sinx or sinx

2 2

Class Exercise Q9.

Solution:

In a ABC , A > B and if A and B satisfy

3 sin x –4 sin3x – k = 0 ( 0< |k| < 1 ) , C is

2 5(a) (b) (c) (d)

3 2 3 6

33sinx 4sin x k 0 sin3A sin3B

3A 3B(A B) A B3

C (A B) C3

2

C3

Class Exercise Q10.

Solution:

Solve the equation

(1-tan)(1+sin2) = 1+tan

(1 tan )(1 sin2 ) 1 tan

22tan

(1 tan ) 1 1 tan1 tan

2

21 tan 2tan

(1 tan ) 1 tan1 tan

Class Exercise Q10.

Solution:

Solve the equation

(1-tan)(1+sin2) = 1+tan

2 2(1 tan ) 1 tan (1 tan )(1 tan )

2(1 tan )[(1 tan )(1 tan ) 1 tan ] 0

2 2(1 tan )(1 tan 1 tan ) 0

2(1 tan )( 2 tan ) 0

2

21 tan 2tan

(1 tan ) 1 tan1 tan

Class Exercise Q10.

Solution:

Solve the equation

(1-tan)(1+sin2) = 1+tan

2when tan 0 m

when 1 tan 0 tan 1

tan tan( )4

n4

m , n , n,m I4

2(1 tan )( 2 tan ) 0