Post on 02-Jun-2018
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Trigonometric Equations : Session 1
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Illustrative Problem
Solve :sinx + cosx = 2
Solution:
no solution.
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A trigonometric equation
is an equation
Contains trigonometric functions of
variable angle
sin =
2 sin2 + sin22 = 2.
Definition
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For sin = , = /6, 5 /6, 13 /6,.
Solution of Trigonometric Equation:
Values of , which satisfy thetrigonometric equation
No. of solutions are infinite .
Why ? - Periodicity of trigonometric functions.
e.g. - sin, cos have a period as 2
Periodicity and general solution
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Periodicityof trigonometric functions.
sinhave a period 2
f(+T) = f()
sin
0 2 3 4
Periodicity and general solution
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Graph of y=sinx
sinx is periodic of period 2
0
3
2
-2
23
2
2
2
-
Y
X
(0,1)
(0,-1)
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Graph of y=cosx
(0,1)
-2 23
2
2
2
3
2
X0
(0,-1)
Y
cosx is periodic of period 2
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tanx is periodic of period
tanx is not defined at x=(2where n is integer
n+1)2
Graph of y=tanx
-2 2322
23
2
X0
Y
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For sin = , = /6, 5/6, 13/6,.
As solutions areinfinite , the entireset of solution can be written in acompactform.
This compact form is referred to as generalsolution
Or = n +(-1)n(/6)General Solution
Periodicity and general solution
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Principal Solutions
Solutions in 0x2
principal solutions.
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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 are5/6 and 11/6
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Illustrative problem
Find the principal solution of theequation sinx = 1/2
Solution
sin/6 = 1/2
and sin(- /6) = 1/2
principal solution are x = /6 and 5/6.
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sin= PM/OP
For sin= 0 , PM = 0
For PM = 0, OP will lie on XOX
Y
X
O
P
M
X
Yis an integral multiple of .
= 0, , 2, 3 ..
General solution of sin = 0
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Or = n , n Z(n belongs to set of integers)
For sin = 0 ,
is an integer multiple of .
Hence, general solution of sin= 0 is
= n , where n Z,
General solution of sin = 0
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cos = OM/OP
For cos = 0 , OM = 0
For OM = 0, OP will lie on YOY
is an odd integer multiple of /2.
= /2, 3/2, 5/2.
Y
X
O
P
M
X
Y
General solution of cos = 0
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Or = (2n+1)/2, n Z (n belongs to set of integers)
For cos = 0 ,
is an odd integer multiple of /2.
Hence, general solution of cos= 0 is
= (2n+1)/2 , where n Z,
General solution of cos = 0
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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 .
= 0,, 2, 3.
Same assin = 0
General solution of tan = 0
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Or = n , n Z(n belongs to set of integers)
For tan= 0 ,
is an integer multiple of .
Hence, general solution of tan= 0 is
= n, where n Z,
General solution of tan = 0
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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
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If sin= k -1k 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
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02
sin2
cos2
02
cos
0
2sin
2)1n2(
2
= (2n+1) -
n2
= 2n +
= n +(-1)n, where n Z
Odd , -veEven , +ve
General solution of sin = k
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If cos= k -1k 1
Let k = cos, choose value of between
0 to
If cos= cos cos- cos= 0
02
sin2
sin2
General solution of cos = k
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02
sin
0
2sin
2
n2
= 2n -
n2
= 2n +
= 2n , where n Z
-ve+ve
02sin2sin2
General solution of cos = k
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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
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2
sin( - ) = 0
- = n , where n Z
= n +
sin.cos- cos.sin= 0
= n+
, where n Z
General solution of tan = k
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Illustrative problem
Find the solution of sinx = 32
Solution:
nx n ( 1)3
As sin x sin3
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Illustrative problem
Solution:
Solve tan2x = cot(x )6
We have tan2x = cot(x )
6
tan( x )2 6
2tan( x)
3
22x n x3
2
x n , where n is an int eger3
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Illustrative problem
Solution:
Solve sin2x + sin4x + sin6x = 0
2sin4x cos2x sin4x 0
sin 4x(2 cos2x 1) 0
1sin4x 0 or cos2x
2
2sin4x 0 or cos2x cos
3
nx or x n , where n is an int eger
4 3
24x n or 2x 2n , where n is an integer3
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Illustrative Problem
Solution:
Solve 2cos2x + 3sinx = 0
22 cos x 3 sinx 0
22(1 sin x) 3 sin x 0
(2 sin x 1)(sin x 2) 0 1
sin x or sin x 22
1 7sin x sin2 6
n7x n ( 1) where n is an int eger6
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General solution of sin2x = sin2cos2x = cos2, tan2x = tan2
n where n is an integer.
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Illustrative Problem
Solve : 4cos
3
x-cosx = 0
Solution:34 cos x cos x 0
2
cos x(4 cos x 1) 0
2cos x 0 or 4 cos x 1
22 21x (2n 1) or cos x cos
2 2 3
x (2n 1) or n / 3, n Z2
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Illustrative Problem
Solve :sinx + siny = 2
Solution:
sin x sin y 2
sin x 1 and sin y 1
sin x sin and sin y sin2 2
n m
x n ( 1) and y m ( 1)2 2
where n and m are int eger.
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Class Exercise Q1.
Solve :sin5x = cos2x
Solution:
cos2x sin5x
cos 2x cos( 5x)2 2x 2n ( 5x)
2
taking positive sign
7x 2n 2
(4n 1)x
14
taking negative sign
2x 2n 5x2
(4n 1)
x , n I6
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Class Exercise Q2.
Solve :2sinx + 3cosx=5
Solution:
2 sin x 3 cos x 5 is possible only when
sin x and cos x attains their max imum
value i.e.sin x 1 and cos x 1.
both sinx and cosx cannot be 1 for any value of x.
hence no solution.
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Class Exercise Q3.
Solve :7cos
2
+3sin
2
= 4
Solution:
2 27 cos 3(1 cos ) 4
2 27 cos 3 3 cos 4
2 1cos4
2 2cos cos3
n , n I3
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Class Exercise Q4.
Solution:
Solve : 2 cos2 2 sin 2
2 cos2 2 sin 2
2 sin 2(cos 2 1) 0 2
2 sin 4 sin 0
3
22 sin (1 2 2 sin ) 0
1sin 0 and sin2
n
n , n ( 1) where n I6
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Class Exercise Q5.
Show that 2cos2(x/2)sin2x = x2+x-2
for 0
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Class Exercise Q6.
Solution:
Find the value(s) of x in (- , )
which satisfy the following equation
2 31 cos x cos x cos x ....to38 4
2 31 cos x cos x cos x ....to 28 8
2 31 cos x cos x cos x ....to 2
12
1 cos x
1cosx
2
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Class Exercise Q6.
Solution:
Find the value(s) of x in (- , ) whichsatisfy the following equation
2 31 cos x cos x cos x ....to38 4
1COSX
2
2x , x3 3
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Class Exercise Q7.
Solve the equation
sinx + cosx = 1+sinxcosx
Solution:
Let sin x cos x z
squaring both sides
21 2 sin x.cos x z 2(z 1)
sinx.cosx2
2z 1sin x cos x 12
2z 2z 1 0 2(z 1) 0 z 1
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Class Exercise Q7.
Solve the equation
sinx + cosx = 1+sinxcosx
Solution:
sin x cos x 1
1 1 1cos x sin x
2 2 2 1cos(x ) cos
4 42
x 2n4 4
(4n )x 2n , , n I
2
2(z 1) 0 z 1
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Class Exercise Q8.
Solution:
3r 4(1 sin x)
sinx
2
2
3 4 sin x 4 sin x
4 sin x 4 sin x 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
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Class Exercise Q8.
If rsinx=3,r=4(1+sinx),
then x is
Solution:
(a) or (b) or 3 4 2
(0 x 2 )
7 5(c) or (d) or
6 6 6 6
(2sinx 1)(2sinx 3) 0
5x or6 6
1 3sin x or sin x
2 2
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Class Exercise Q9.
Solution:
In a ABC , A > B and if A
and
B satisfy3 sin x 4 sin3x k = 0 ( 0< |k|< 1 ) , C is
2 5(a) (b) (c) (d)
3 2 3 6
33 sin x 4 sin x k 0 sin3A sin3B
3A 3B(A B) A B 3
C (A B) C3
2C
3
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Class Exercise Q10.
Solution:
Solve the equation(1-tan)(1+sin2) = 1+tan
(1 tan )(1 sin2 ) 1 tan
2
2tan(1 tan ) 1 1 tan
1 tan
2
2
1 tan 2 tan(1 tan ) 1 tan
1 tan
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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
2
1 tan 2 tan(1 tan ) 1 tan
1 tan
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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