trigonometric equations : session 1 illustrative problem solve :sinx + cosx = 2 solution:
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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