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Trigonometric Equations : Session 1
Illustrative Problem
Solve :sinx + cosx = 2
Solution:
no solution.
Definition
A trigonometric equation
 is an equation
 Contains trigonometric functions of
variable angle
sin  = ½
2 sin2 + sin22 = 2.
Periodicity and general solution
Solution of Trigonometric Equation:
Values of , which satisfy the
trigonometric equation
For sin  = ½ ,  =  /6, 5  /6, 13  /6,…….
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.
f(+T) = f()
sin
sin have a period 2 
0

2
3
4

Graph of y=sinx
sinx is periodic of period 2 
Y
(0,1)
0
-2  3 -  
2
2
(0,-1)

2

3
2
2
X
Graph of y=cosx
cosx is periodic of period 2
Y
(0,1)
0
-2  3
2

2
(0,-1)


2

3
2
2
X
Graph of y=tanx
tanx is periodic of period 

tanx is not defined at x=(2n+1)
2
where n is integer
Y
0
-2  3
2


2

2

3
2
2
X
Periodicity and general solution
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
For sin  = ½ ,  = /6, 5/6, 13/6,…….
Or  = n  +(-1)n ( /6)
General Solution
Principal Solutions
Solutions in 0x2
 principal solutions.
Illustrative problem
Find the principal solutions of
tanx =  1
3
Solution
We know that tan( - /6) = 
and tan(2 - /6) =

1
1
3
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.
General solution of sin  = 0
sin = PM/OP
For sin = 0 , PM = 0
Y
P
For PM = 0, OP will lie on XOX’
  = 0, ±π, ±2π, ±3π …..
  is an integral multiple of π.
X

X’
O
M
Y’
General solution of sin  = 0
For sin  = 0 ,
 is an integer multiple of π.
Or  = nπ , n є Z (n belongs to set of integers)
Hence, general solution of sin = 0 is
 = nπ , where n є Z,
General solution of cos  = 0
cos  = OM/OP
For cos  = 0 , OM = 0
Y
For OM = 0, OP will lie on YOY’
  = ±π/2, ±3π/2, ±5π/2….
P
O
 is an odd integer multiple of π/2.
X

X’
M
Y’
General solution of cos  = 0
For cos  = 0 ,
 is an odd integer multiple of π/2.
Or  = (2n+1)π/2 , n є Z (n belongs to set of integers)
Hence, general solution of cos = 0 is
 = (2n+1)π/2 , where n є Z,
General solution of tan  = 0
tan = PM/OM
For tan = 0 , PM = 0
Y
For PM = 0, OP will lie on XOX’
P
  = 0,±π, ±2π, ±3π….
 is an integer multiple of π.
Same as
sin  = 0
X

X’
O
M
Y’
General solution of tan  = 0
For tan  = 0 ,
 is an integer multiple of π.
Or  = nπ , n є Z (n belongs to set of integers)
Hence, general solution of tan = 0 is
 = nπ, where n є Z,
Illustrative Problem
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
General solution of sin = k
If sin = k
 -1 k  1
Let k = sin, choose value of 
between –/2 to  /2
If sin = sin  sin - sin = 0




  



 sin
 0
2cos




2 
2 







General solution of sin = k
2cos     sin     0


2 
2 






cos    0
2

















    (2n 1) 
2
2





 = (2n+1)π - 





 0
sin
2 











     n
2 
 = 2nπ + 
Odd , -ve
 = nπ +(-1)n , where n є Z
Even , +ve
General solution of cos = k
If cos = k
 -1 k  1
Let k = cos, choose value of  between
0 to 
If cos = cos  cos - cos = 0




  



 sin
 0
2sin




2 
2 







General solution of cos = k
2sin     sin     0


2 
2 





sin    0
2







2











     n
2 
 = 2nπ - 






 0
sin
2 


  

  n

2 







 = 2nπ + 
-ve
 = 2nπ   , where n є Z
+ve
General solution of tan = 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
sin.cos - cos.sin = 0
 sin(  -  ) = 0

2

 -  = nπ , where n є Z

 = nπ + 
 = nπ+  , where n є Z
Illustrative problem
Find the solution of sinx =
3
2
Solution:
As sin x  sin

3
 x  n  (1)n

3
Illustrative problem

Solve tan2x =  cot(x  )
6
Solution:

)
6


 tan(  x  )
2
6
We have tan2x =  cot(x 
2
 2x  n 
x
3
 x  n 
2
 tan(
 x)
3
2
, where n is an int eger
3
Illustrative problem
Solve sin2x + sin4x + sin6x = 0
Solution:
2 sin 4x cos 2x  sin 4x  0
 sin4x(2 cos2x  1)  0
 sin 4x  0 or cos 2x  
1
2
 sin 4x  0 or cos2x  cos
2
3
2
, where n is an integer
3
n

x 
or x  n  , where n is an int eger
4
3
 4x  n or 2x  2n 
Illustrative Problem
Solve
2cos2x + 3sinx = 0
Solution:
2 cos2 x  3 sin x  0
 2(1  sin2 x)  3 sin x  0
 (2 sinx  1)(sinx  2)  0
 sin x  
1
 sin x  
or sin x  2
2
1
7
 sin
2
6
7
 x  n  (1)
where n is an int eger
6
n
General solution of sin2x = sin2
cos2x = cos2, tan2x = tan2
n   where n is an integer.
Illustrative Problem
Solve : 4cos3x-cosx = 0
Solution:
4 cos3 x  cos x  0
 cos x(4 cos2 x  1)  0
 cos x  0 or 4 cos2 x  1
2


1
 x  (2n  1) or cos2 x     cos2
2
3
2

 x  (2n  1) or n   / 3, n  Z
2
Illustrative Problem
Solve :sinx + siny = 2
Solution:
sinx  siny  2
 sinx  1 and siny  1


and sin y  sin
2
2
n 
m 
 x  n  (1)
and y  m  (1)
2
2
 sin x  sin
where n and m are int eger.
Class Exercise Q1.
Solve :sin5x = cos2x
Solution:
cos 2x  sin5x

 cos 2x  cos(  5x)
2
taking positive sign

 7x  2n 
2
(4n  1)
x
14

 2x  2n  (  5x)
2
taking negative sign

2x  2n   5x
2
(4n  1)
x
, nI
6
Class Exercise Q2.
Solve :2sinx + 3cosx=5
Solution:
2 sinx  3cos x  5 is possible only when
sinx and cos x attains their maximum
value i.e.sinx  1 and cos x  1.
both sinx and cos x cannot be 1 for any value of x.
hence no solution.
Class Exercise Q3.
Solve :7cos2 +3sin2 = 4
Solution:
2
2
7 cos   3(1  cos )  4
2
2
 7 cos   3  3 cos   4
 cos2  
1
4
 cos2   cos2

3

   n  , n  I
3
Class Exercise Q4.
Solve : 2 cos2  2 sin   2
Solution:
2 cos2  2 sin   2
 2 sin   2(cos2  1)  0
2
 2 sin   4 sin   0
3
2
 2 sin (1  2 2 sin )  0
1
 sin   0 and sin  
2
n 
   n , 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:
x
2
L.H.S.  2 cos   sin x
2
2
2
 (1  cos x).sin x  2
2
R.H.S  x 
1
2
x2
1

 x    2  2
x

Hence no solution
Class Exercise Q6.
Find the value(s) of x in (- , )
which satisfy the following equation
1 cos x  cos2 x  cos3 x ....to 
8
 43
Solution:
1 cos x  cos2 x  cos3 x ....to 
8
 82
 1  cos x  cos2 x  cos3 x  ....to   2

1
2
1  cos x
 cos x 
1
2
Class Exercise Q6.
Find the value(s) of x in (- , )
satisfy the following equation
1 cos x  cos2 x  cos3 x ....to 
8
Solution:
1
 COSX  
2
x  

2
,x  
3
3
which
 43
Class Exercise Q7.
Solve the equation
sinx + cosx = 1+sinxcosx
Solution:
Let sinx  cos x  z
squaring both sides
(z2  1)
1  2 sin x.cos x  z
 sin x.cos x 
2
 z2  1 
 sin x  cos x  1  

 2 


2
 z2  2z  1  0
 (z  1)2  0  z  1
Class Exercise Q7.
Solve the equation
sinx + cosx = 1+sinxcosx
Solution:
(z  1)2  0  z  1
 sin x  cos x  1

1
2
cos x 
 x  2n,
1
2
sin x 
1
2
(4n)
, nI
2

1

 cos(x  ) 
 cos
4
4
2
 
 x  2n  
4 4
Class Exercise Q8.
If rsinx=3,r=4(1+sinx),
then x (0  x  2) is



(a) or
(b) or 
3 4
2
 7
 5
(c) or
(d) or
6
6
6
6
Solution:
3
r
 4(1  sin x)
sin x
 3  4 sin x  4 sin2 x
 4 sin2 x  4 sin x  3  0
 (2 sinx  1)(2 sinx  3)  0
Class Exercise Q8.
If rsinx=3,r=4(1+sinx),
then x (0  x  2) is


or
3
4

7
(c) or
6
6
(a)

or 
2

5
(d)
or
6
6
(b)
Solution:
 (2 sinx  1)(2 sinx  3)  0
 sin x 
1
3
or sin x  
2
2
x
 5
or
6
6
Class Exercise Q9.
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
Solution:
3 sin x  4 sin3 x  k  0
 3A    3B(A  B)
 C    (A  B)
2
C 
3
 sin3A  sin3B

 A B 
3

C 
3
Class Exercise Q10.
Solve the equation
(1-tan)(1+sin2) = 1+tan
Solution:
(1  tan )(1  sin2)  1  tan 

2 tan  
 (1  tan ) 1 
 1  tan 
2 
1  tan  

 1  tan2   2 tan  
 (1  tan ) 
  1  tan 
2


1  tan 


Class Exercise Q10.
Solve the equation
(1-tan)(1+sin2) = 1+tan
Solution:
 1  tan2   2 tan  
 (1  tan ) 
  1  tan 
2


1  tan 


2
 (1  tan ) 1  tan    (1  tan )(1  tan2 )
 (1  tan )[(1  tan )(1  tan )  1  tan2 ]  0
 (1  tan )(1  tan2   1  tan2 )  0
 (1  tan )(2 tan2 )  0
Class Exercise Q10.
Solve the equation
(1-tan)(1+sin2) = 1+tan
Solution:
 (1  tan )(2 tan2 )  0
when tan2   0    m
when 1  tan   0  tan   1

 tan   tan( )
4

   n 
4

   m, n  , n,m  I
4