09. Trigonometric Equations-1

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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