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Math Help - Trig equation

  1. #1
    Senior Member Stroodle's Avatar
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    Trig equation

    Hi there,

    I'm having a little trouble solving the following equation for -\pi <x\leq 2 \pi:

    sin^2(x)cos^3(x)=cos(x)

    I've rearranged it into the form:

    cos(x)\left [sin^2(x)cos^2(x)-1\right ]=0

    So where cos(x)=0,\ \ \ x= \pm\frac{\pi}{2} and \frac{3\pi}{2}

    But I'm just not sure how to find x where sin^2(x)cos^2(x)=1

    Thanks for your help.
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by Stroodle View Post
    Hi there,

    I'm having a little trouble solving the following equation for -\pi <x\leq 2 \pi:

    sin^2(x)cos^3(x)=cos(x)

    I've rearranged it into the form:

    cos(x)\left [sin^2(x)cos^2(x)-1\right ]=0

    So where cos(x)=0,\ \ \ x= \pm\frac{\pi}{2} and \frac{3\pi}{2}

    But I'm just not sure how to find x where sin^2(x)cos^2(x)=1

    Thanks for your help.
    \sin^2{x}(1-\sin^2{x}) = 1

    \sin^2{x} - \sin^4{x} = 1

    0 = \sin^4{x} - \sin^2{x} + 1

    let u = \sin^2{x} ...

    0 = u^2 - u + 1

    note that b^2-4ac < 0 ... therefore, this factor of your original equation has no solution.
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  3. #3
    Senior Member Stroodle's Avatar
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    ahh, of course

    Thanks!
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  4. #4
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    Hello, Stroodle!

    Another approach . . .


    Solvie the following equation for -\pi < x\leq 2 \pi\!:\;\;\sin^2\!x\cos^3\!x\:=\L\cos x

    I've rearranged it into the form: \cos(x)\left[\sin^2\!x\cos^2\!x-1\right ]\:=\:0

    So: \cos x\,=\,0,\;x\:=\: \pm\frac{\pi}{2}\,\text{ and }\,\frac{3\pi}{2}

    But I'm not sure how to solve: \sin^2x\cos^2x \:=\:1

    Take square roots: . \sin x\cos x \:=\:\pm1

    \text{Multiply by 2: }\;\underbrace{2\sin x\cos x} \:=\:2

    \text{Then we have: }\quad\;\;\overbrace{\sin2x} \:=\:2 .[1]


    But |\sin2x| \,\leq\,1


    . . Therefore, [1] has no real solutions.

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