Download Session Objectives

Mathematics
Trigonometric Equations – Session 2
Session Objectives
Session Objectives
Removing Extraneous Roots
Avoiding Root Loss
Equation of the form of a cosx +b sinx = c
Simultaneous equation
Trigonometric Equation –
Removing Extraneous Solutions
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Extraneous solutions
The solutions, which do not satisfy
the trigonometric equation.
Origin of Extraneous solutions ?
Squaring during solving the trigo. equations. ( ±)2= +
Solutions, which makes the equation undefined.
( denominator being zero for a set of solutions)
Esp . equations containing tan or sec, cot or cosec
Trigonometric Equation –
Extra root because of squaring
Illustrative problem
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Solve sec - 1 = ( √ 2 – 1) tan
Equation can be rewritten as
sec = ( √ 2 – 1) tan + 1
On squaring , we get
sec2 = (2+1–2√2)tan2 +1+2(√2–1)tan
sec2 - tan2 = (2–2√2)tan2 +1 + 2(√2–1)tan
(2 – 2√2 ) tan2 + 2(√2 – 1) tan = 0
tan  = 0 and tan  = +1
Trigonometric Equation –
Extra root because of squaring
Solve sec - 1 = ( √ 2 – 1) tan
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tan  = 0 or tan  = +1
 = n or  = n + /4
Putting  =  in given equation , we get
L.H.S. = ( -1 –1) = -2
R.H.S. = ( √ 2 – 1).0 = 0
Extraneous solutions :  = odd integer multiple of π
WHY ??
Answer : = 2n or  = 2n + /4
Trigonometric Equation –
Solutions, which makes the
equation undefined
Solve
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tan5 = tan3
Solution will be 5 = nπ + 3
  = nπ/2 , where n  Z
solutions  = π/2, 3π/2…..etc. will make tan3 and
tan5 undefined
Answer :  = mπ, where m  Z
Trigonometric Equation –
Avoiding root loss
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Reason for root loss ?
 Canceling the terms from
both the sides of the equation
 Use of trig. Relationship , which restricts the
acceptable values of  ,i.e., the domain of  changes.



2
1 tan 
2 
cos 
1 tan2  
2










Trigonometric Equation –
Cancelling of terms from both sides
Illustrative problem
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Solve sin .cos = sin
If we cancel sin  from both the sides
cos  = 1   = 2n , where n  Z
However , missed the solution provided by sin = 0
  = n ,where n  Z
Answer :  = n ,where n  Z
Trigonometric Equation –
Changes in domain of 
Illustrative Problem
Solve : sin - 2.cos = 2
If we use
1 tan2 
2
cos 
1 tan2 
2























2tan 
2 
sin  
1 tan2  
2










Equation can be written as :









1 tan2

2tan


2   2.
2   2

2  
1 tan2  
1 tan

2
2 











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Trigonometric Equation –
Changes in domain of 
Solve :









2 tan
2









sin - 2.cos = 2









 2.
1  tan2
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 







2 2
1  tan2 
1  tan2 
2
2
2 tan   2  2 tan2   2  2 tan2 
2
2
2
 tan   2   = 2nπ + 2 , where n  Z and  = tan-1 2
2
Root loss :  = (2n+ 1) π where , n  Z
As  = π , 3 π , 5 π …. satisfy the given trig.
equation
Answer :  = (2nπ+2) U (2n+1)π, where n  Z and
= tan-1 2
Trigonometric Equation –
a sin  + b cos  = c
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Reformat the equation in the form of
cos(  -  ) = k.
1. Divide both sides of the equation
by
a2  b2
2. The equation now will be as
a
b
c
sin 
cos 
a2  b2
a2  b2
a2  b2
Trigonometric Equation –
a sin  + b cos  = c
a
b
c
sin 
cos  
a2  b2
a2  b2
a2  b2
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Compare with sin.sin + cos.cos = k
sin  
a
a2  b2
cos  
b
a2  b2
Hence , the given equation can be written as
a
c

1
cos (  ) 
, where   tan  
b
a2  b2
For real values of  , 1 
c
1
a2  b2
a2  b2

b
a
Trigonometric Equation –
a sin  + b cos  = c - Algorithm
Step 1:
Reformat the equation into
cos(-) = k
Step 2:
Check whether real solution exists
1 
c
1
a2  b2
Step 3:
Solve the equation cos(-) = k
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Trigonometric Equation –
a sin  + b cos  = c - Problem
Illustrative Problem
Solve sin + cos = 1
Step 1:
Reformat the equation into cos(-) = k
1
1
1
sin 
cos  
12 12
12 12
12 12
1 sin  1 cos  1
2
2
2


cos    1
4
2





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Trigonometric Equation –
a sin  + b cos  = c - Problem
Solve sin + cos = 1
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

cos     1
4

2
Step 2:
Check whether real solution exists
c
1 
1
a2  b2
As
1 is in 1,1




2
Real solution exists
Trigonometric Equation –
a sin  + b cos  = c - Problem
Solve sin + cos = 1
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

cos     1
4

2
Step 3:
Solve the equation cos(-) = k


cos    cos 
4
4





   2n  
4
4
 2n or
 2n  
2
Simultaneous Trigonometric
Equations
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Case I : Two equations and one
variable angle
Step 1 – Solve both the equation between 0 and 2π .
Step 2 – Find common solutions.
Step 3– Generalise the solution by adding 2nπ to
common solution as per step 2.
Simultaneous
Trigonometric
Equations -
Problem
Find the general solution of
tan = -1, cos = 1/2
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Step 1 – Solve both the equation between 0 and 2π
For tan  1,   3,7
4
4
For cos  1 ,   ,7
4 4
2
Simultaneous Trigonometric
Equations - Problem
Illustrative Problem
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Find the general solution of
tan = -1, cos = 1/2
Step 2 – Find common solutions

7
4
Step 3– Generalise the solution
  2n 
7
4
Answer
Simultaneous Trigonometric
Equations
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Case II :
Two equations and
two variable angles(θ,φ) and
smallest positive values of the
angles satisfying the equations
needs to be found out
Find the smallest positive values of θ and φ ,
if tan(θ – φ) = 1 and sec ( θ+φ) = 2/ √3
Simultaneous Trigonometric
Equations
Step 1 – Solve both the equations
between 0 and 2π .
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Step 2 –
Find two equations with the conditions such as
( θ+φ) > (θ-φ) for positive θ and φ
Step 3 - Solve the two equations to determine θ and
φ
Simultaneous Trigonometric
Equations - Problem
Illustrative Problem
Find the smallest positive values of θ and φ
, if tan(θ – φ) = 1 and sec ( θ+φ) = 2/ √3
Step 1 –
Solve both the equations
between 0 and 2π .
  
  
 5
,
4 4
 11
,
6 6
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Simultaneous Trigonometric
Equations - Problem
Find the smallest positive values of θ and φ
, if tan(θ – φ) = 1 and sec ( θ+φ) = 2/ √3
 5
   ,
4 4
  
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 11
,
6 6
Step 2 – Find two equations with the conditions such as ( θ+φ) >
(θ-φ) for positive θ and φ
11
  
6

  
4
Step 3 - Solve the two
equations to determine θ and
φ
25
19

;
24
24
Trigonometric Equation –
Misc. Tips :
1. cosθ = k is simpler to solve
compared to sinθ = k
2. Check for extraneous roots and
root loss
3. In case of ‘algebraic function of angle’ is a part of the
equation use of the following properties:
•
Range of values of sin and cos
functions
•
x2 ≥ 0
•
A.M. ≥ G.M.
2

x
 x 
2
2 cos
 2 x  2x
 6 


Trigonometric Equation –
Misc. Tips :
4. Equation with multiple terms of
the form (sinθ ± cosθ) and
sinθ.cosθ :
•
Put sinθ+cosθ = t.
•
Equation gets converted in to a quadratic equation
in t
sin x + cos x = 1+ sin x. cos x
Trigonometric Equation –
Misc. Tips :
5. Equation with terms of sin2θ ,
cos2θ and sinθ.cosθ
•
Try dividing by cos2θ to get a
quadratic equation in tan θ.
Solve the equation 2 sin2 x – 5 sinx.cos x – 8 cos2 x = -2
6. Solution to sin2θ = sin2 , cos2θ = cos2  and tan2θ
= tan2  is n  
Class exercise Q1
Number of solution of the equation
tanx+secx = 2cosx lying in the
interval [0,2]
(a) 2
(b) 3
(c ) 0
Solution:
tanx  sec x  2 cos x wherex  [0,2]
 sin x  1  2 cos2 x
 2  2 sin2 x  sin x  1
cos x  0
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(d) 1
Class exercise Q1
Number of solution of the equation
tanx+secx = 2cosx lying in the
interval [0,2]
 2 sin2 x  sin x  1  0
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 2 sin2 x  2 sin x  sin x  1  0
 2 sinx(sinx  1)  1(sinx  1)  0  sin x 
1
or sin x  1
2


n
x  n  (1)
or x  n  (1) ( )
6
2
n
for x  [0,2]


n0 x
or x   not valid
6
2
Class exercise Q1
Number of solution of the equation
tanx+secx = 2cosx lying in the
interval [0,2]
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5
3
13
3
n1 x 
or x 
n2x
( 2) or x 
6
2
6
2
 5 3
x  ,
,
6 6 2
3
however for x 
, tan x and sec x is undefined.
2
 5
hence solution in [0,2] is x  ,
6 6
Class exercise Q2
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If 2 sec x  tanx  1, x is



(a)n  (b)n 
(c)2n 
4
4
4
Solution:
c none
 2 sec x  (1  tanx)
 2 sec2 x  1  tan2 x  2 tan x
 2  2 tan2 x  1  tan2x  2 tan x
 tan2 x  2 tan x  1  0

 x  n 
4
 (tan x  1)2  0  tan x  1
Class exercise Q2
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If 2 sec x  tanx  1, x is
As we have squared both the sides,
we should check for extraneous roots
2


L.H.S

2.
 1  2  1  1  R.H.S
n  0, x  
1
4
3
 L.H.S  2. 2  1  2  1  3  R.H.S
n  1, x  
4
 
Similarly, for n=1, 3, 5 …the values of x do not
satisfy the question.

Hence, the solution is x  2n 
4
Class exercise Q3
For nz, the general solution of
the equation


3  1 sin x 


3  1 cos x  2 is


(a) x  n   1

4 12

n 
(c) x  2n   1

4 12
n


(b) x  2n  
4 12


(d) x  2n  
4 12
Solution:


3  1 sin x 
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

3  1 cos x  2
Class exercise Q3
_J32
For nz, the general solution of
the equation


3  1 sin x 




3  1 sin x
2 2

3  1 cos x  2 is


2 2



 cos  x 
 cos



12
4


 x  2n  
4 12
 cos x 
3 1
2
2 2
Class exercise Q4
1
sin(x  y)  cos(x  y)  ,
2
the values of x and y lying between
0o and 90o are given by
(a) x=15o , y=25o (b) x=65o , y=15o
(c) x=45o , y=45o (d) x=45o , y=15o
Solution:
1
if sin(x  y) 
2
 5
xy ,
6 6
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Class exercise Q4
1
sin(x  y)  cos(x  y)  ,
2
the values of x and y lying between
0o and 90o are given by
 5
1
xy '
cos (x  y) 
3 3
2

 
for x and y   0,   x  y 
6
 2



(x  y) 
3  2x  2  x  4  45

 x  y   y  30
6
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Class exercise Q5
Solve (sin x  cos x)2  (sin x  1)2
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(cos x  1)2  0
Solution:
L.H.S =0 if sinx-cosx=0, sinx-1=0 cosx-1=0
 sinx=cosx, sinx=1, cosx=1
Which is not possible for any value of x.
 No Solution
Class exercise Q6
If cot x  cos ecx  3, then x is


(a) 2n 
(b) n 
3
3

(c) n 
(d) None of these
3
Solution:
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given cot x  cos ecx  3
cos ecx 

3  cot x

cos ec2x  3  cot2 x  2 3 cot x (squaring)
 2  2 3 cot x  0
Class exercise Q6
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If cot x  cos ecx  3, then x is
cot x 
1
3

 x  n 
3


for x    ,    .....LHS  RHS
3
3

Hence acceptable solution is x  2n 
3
Class exercise Q7
Solve sinx  3 cos x  2
Solution:
the given equation of the form a
cos + b sin = c
Here a  3,b  1,c  2
for real solution
| c |
a2  b2
i.e| 2 | 3  1
i.e| 2 | 2 whish is true
_J32
Class exercise Q7
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Solve sinx  3 cos x  2
Hence dividing both side by
a2  b2 i.e. 2, we get
3
1
1
cos x  sin x 
2
2
2



 cos x.cos  sin x.sin  cos
6
6
4



 cos  x    cos

6
4
Class exercise Q7
_J32
Solve sinx  3 cos x  2


 x   2n 
6
4
Taking positive sign


x   2n 
6
4
 
5
 x  2n    2n 
4 6
12
Class exercise Q7
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Solve sinx  3 cos x  2
Taking Negative sign


x   2n 
6
4
 

 x  2n    2n 
4 6
12
5

 x  2n 
, 2n 
where n  I
12
12
Class exercise Q8
Solve the equation
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2 sin2 x  5 sin x cos x  8 cos2 x  2
Solution:
In such types of problems we divide
both sides by cos2x which yield a
quadratic equation in tanx.

povided cos x  0 i.e. x 
2
In this equation if cosx = 0, the
equation becomes 2sin2x=-2 or sin2x=-1
which is not possible
hence on dividing the equation by cos2x we get
Class exercise Q8
Solve the equation
2 sin2 x  5 sin x cos x  8 cos2 x  2
2tan2x-5tanx-8 = -2sec2x
2tan2x+2(1+tan2x)-5tanx-8 = 0
or 4tan2x-5tanx-6 = 0
or 4z2-5z-6 = 0 where z = tanx
or 4z2-8z+3z-6 = 0
 4z(z-2)+3(z-2)=0  z=2,-3/4
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Class exercise Q8
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Solve the equation
2 sin2 x  5 sin x cos x  8 cos2 x  2
3
i.e tan x  2; tan x 
4
1
 x  tan
1  3 
2; x  tan
1
 x  n  tan
 4 

1  3  
2; x  n  tan    ,n  I
 4 

Class exercise Q9
Determine for which value of ‘a’ the
equation a2– 2a+sec2(a+x)=0 has
solution and find the solution
The equation involves two unknown a
and x so we must get two condition for
determining unknowns since R.H.S is
zero. So break the L.H.S of the
equation as sum of two square.
a2  2a  sec2 (a  x)  0
or a2  2a  1  tan2 (a  x)  0
_J33
Class exercise Q9
Determine for which value of ‘a’ the
equation a2– 2a+sec2(a+x)=0 has
solution and find the solution
_J33
 (a  1)2  tan2 (a  x)  0
 (a  1)2  0, tan2 (a  x)  0
 a=1 and tan(a+x)=0 but a=1
 (1  x)  n
 x  n    (n  1)
 x  n  1 where n  I
 x  m where m  I
Class exercise Q10
Solve cot – tan = sec 
cos  sin 
1
cos 2
1




sin  cos  cos 
sin  cos  cos 
 1  2 sin2   sin  (cos   0)
 2 sin2   sin   1  0
 (2 sin   1)(sin   1)  0
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Class exercise Q10
Solve cot – tan =sec 
1
n 
 sin      n  (1)
2
6
But sin   1  sec  is undefined
 sin   1
_J30