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Math Help - Find the value of k:

  1. #1
    Member great_math's Avatar
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    Find the value of k:

    Find the values of k for which the equation

    \sin^2 x-|k-2|\sin x=\cos^2 x+|k-2|-1

    is true
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  2. #2
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    Quote Originally Posted by great_math View Post
    Find the values of k for which the equation

    \sin^2 x-|k-2|\sin x=\cos^2 x+|k-2|-1

    is true
    Hi

    You can first get a quadratic in \sin x by substituting \cos^2 x - 1 with -\sin^2 x
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by great_math View Post
    Find the values of k for which the equation

    \sin^2 x-|k-2|\sin x=\cos^2 x+|k-2|-1

    is true

    sin^2x - \mid k-2 \mid sinx = cos^2 x + \mid k-2 \mid -1

    sin^2 x - \mid k-2 \mid sinx = -sin^2x + \mid k-2 \mid

    2sin^2x = (\mid k-2 \mid) (sinx +1 )

    \frac{2sin^2x }{sinx +1 } = \mid k-2 \mid

    k-2 = \frac{2sin^2x }{sinx +1 } \Rightarrow k=\frac{2sin^2x }{sinx +1 }+2

    k-2 = \frac{-(2sin^2x) }{sinx +1 }\Rightarrow k=\frac{-(2sin^2x) }{sinx +1 }+2

    x \ne \frac{3\pi}{2} + 2n\pi since if you do not put this condition the denominator will equal zero and k will be infinity so


    k=\frac{2sin^2x }{sinx +1 }+2 \Rightarrow 2\leq k < \infty

    k=\frac{-(2sin^2x) }{sinx +1 }+2 \Rightarrow -\infty< k \leq 2

    so

     -\infty < k < \infty but I am not sure
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Amer View Post
    sin^2x - \mid k-2 \mid sinx = cos^2 x + \mid k-2 \mid -1

    sin^2 x - \mid k-2 \mid sinx = -sin^2x + \mid k-2 \mid

    2sin^2x = (\mid k-2 \mid) (sinx +1 )

    or

    2sin^2x = \mid k-2 \mid (sin(x))+\mid k-2 \mid

    2sin^2x -\mid k-2 \mid (sin(x))-\mid k-2 \mid  =0

    let u=sinx

    2u^2 -\mid k-2 \mid (u) -\mid k-2 \mid  =0

    u=\frac{\mid k-2 \mid  \mp \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}}{4}

    you know that -1\leq sinx\leq 1

    -1\leq \frac{\mid k-2 \mid - \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}}{4}\leq 1

    -1\leq \frac{\mid k-2 \mid + \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}}{4}\leq 1


    -4\leq (\mid k-2 \mid - \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )})\leq 4


    -4\leq (\mid k-2 \mid + \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )})\leq 4


    0\leq [\mid k-2 \mid + \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}]\times [\mid k-2 \mid - \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}]\leq 16

    0\leq (\mid k-2 \mid)^2 - [(\mid k-2 \mid)^2 -8\mid k-2 \mid] \leq 16

    0 \leq 8 \mid k-2 \mid \leq 16

     0 \leq \mid k-2 \mid \leq 2

     \mid k-2 \mid \leq 2

     -2\leq k-2 \leq 2

    0 \leq k \leq 4

    hmm
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  5. #5
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    Hello, great_math!

    A puzzling problem . . . What is meant by "is true"?


    Find the values of k for which the equation: . \sin^2\!x-|k-2|\sin x=\cos^2\!x+|k-2|-1 .is true.
    Replace \cos^2\!x with 1 - \sin^2\!x:

    . . \sin^2\!x - |k-2|\sin x \;=\;1 - \sin^2\!x +|k-2| - 1

    . . 2\sin^2\!x - |k-2|\sin x - |k-2| \;=\;0


    Quadratic Formula: . \sin x \;=\;\frac{|k-2| \pm\sqrt{|k-2|^2 + 8|k-2|}}{4}



    Now it all depends on what they're asking for . . .


    . . The discriminant must be nonnegative: . |k-2|^2 + 8|k-2| \;\geq \;0

    . . Since |\sin x| \:\leq \:1\!:\quad \left|\frac{|k-2| \pm\sqrt{|k-2|^2 + 8|k-2|}}{4}\right| \;\leq \;1


    Good luck!

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  6. #6
    Senior Member pankaj's Avatar
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    Putting t=\sin\ x.What we have is the quadratic

    2t^2-|k-2|t-|k-2|=0 where t\in[-1,1]

    Discriminant = |k-2|^2+8|k-2|\geq 0

    Thus we will get real values of t for all real values of k.

    But we are required to ensure that these values must lie on the interval [-1,1]

    Let f(t)=2t^2-|k-2|t-|k-2|=0

    CASE 1

    If exactly one root of f(t)=0 lies on (-1,1),then we must have

    f(-1)f(1)<0

    Now, f(-1)=2>0

    \therefore f(1)<0,thus 2-2|k-2|<0

    |k-2|>1

    k-2<-1 or k-2>1

    k<1 or k>3i.e. k\in(-\infty,1)\cup(3,\infty)

    CASE 2

    If both roots on the interval (-1,1) then following three conditions must be satisfied:
    (1) f(-1)>0

    (2) f(1)>0

    (3) -1<\frac{\alpha+\beta}{2}<1; \alpha and \beta being the rootsof the equation f(t)=0

    f(-1)=2>0

    f(1)>0 implies that 2-2|k-2|>0.

    \therefore |k-2|<1

    Thus -1<k-2<1.

    k\in (1,3)

    -1<\frac{\alpha+\beta}{2}<1 implies that -1<\frac{|k-2|}{4}<1

    (Recall that if \alpha and \beta are roots of ax^2+bx+c=0 then \alpha+\beta=-\frac{b}{a})

    On solving for k we get -2<k<6

    Taking intersection of solutions of (1),(2) and (3),we get 1<k<3


    Finally taking union of Case 1 and Case 2 and also checking for k=1,3, we arrive at the conclusion that the given equation is true for all real k.

     <br />
\therefore k\in R<br />
must be the answer.
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