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Math Help - Trigonometric equation involving (sin x ) ^ 2

  1. #1
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    Trigonometric equation involving (sin x ) ^ 2

    So this is the equation,
    (sin x)^2 = sinx cosx for x lying between 0 and 360 degrees both inclusive.

    Solving it, i got the following values for x

    x = 0, 45 , 135 , 180 , 225 , 315 , 360

    but the correct values for x are x = 0, 45 , 180 , 225 , 360

    I dont understand why 135, 315 is not considered as correct .
    Nid ur help guys. thnks in advance
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Trigonometric equation involving (sin x ) ^ 2

    We are given to solve:

    \sin^2(x)=\sin(x)\cos(x) where 0^{\circ}\le x\le360^{\circ}

    \sin^2(x)-\sin(x)\cos(x)=0

    \sin(x)(\sin(x)-\cos(x))=0

    Now, in turn equate both factors to zero, and solve for x on the given interval:

    \sin(x)=0

    x=0^{\circ},180^{\circ},360^{\circ}

    \sin(x)-\cos(x)=0

    \sin(x)=\cos(x)

    x=45^{\circ},225^{\circ}

    The extra solutions you gave satisfy \sin(x)+\cos(x)=0, which is not part of the original equation.
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  3. #3
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    Re: Trigonometric equation involving (sin x ) ^ 2

    sin x = cos x
    i solved it this way..

    squaring both sides,
    (sin x) = (cos x)
    (sin x) = 1 - (sin x)
    2 (sin x) = 1
    sin x = (1/√2)

    i get x= 45,135 and x=225,315

    how do i reject 135 and 315 ?
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  4. #4
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    Re: Trigonometric equation involving (sin x ) ^ 2

    how do i reject 135 and 315 ?
    you're expected to know that sine and cosine have opposite signs at x = 135 and x = 315
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  5. #5
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    Re: Trigonometric equation involving (sin x ) ^ 2

    Yeah ..should be more careful about these mistakes next time . Thank you dude : ))
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  6. #6
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Trigonometric equation involving (sin x ) ^ 2

    x= \left\{0, \frac{\pi }{4}, \pi ,\frac{5\pi }{4}\right\}
    Last edited by MaxJasper; October 11th 2012 at 12:13 PM.
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