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Math Help - Just want to verify I am doing this right

  1. #1
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    Just want to verify I am doing this right

    I did get the right answers, but I still would like to make sure my thinking is accurate, and also check if I made unnecessary calculations somewhere.

    Get all the solutions to the trigonometric equation \sin^4 x + \cos^4 x = 1 in the interval [0, 2pi[

    What I did was:

    \sin x\sin x\sin x\sin x + \cos x\cos x\cos x\cos x = 1

    <br />
[-\dfrac{1}{2}(\cos 2x - \cos 0)][-\dfrac{1}{2}(\cos 2x - \cos 0)] + [\dfrac{1}{2}(\cos 2x + \cos 0)][\dfrac{1}{2}(\cos 2x + \cos 0)] = 1

    <br />
[-\dfrac{1}{2}(2\cos^2 x - 2)][-\dfrac{1}{2}(2\cos^2 x - 2)] + [\dfrac{1}{2}(2\cos^2 x)][\dfrac{1}{2}(2\cos^2 x)] = 1

    (-\cos^2 x - 1)^2 + \cos^4 x = 1

    2\cos^4 x - 2\cos^2 x + 1 = 1

    \cos^4 x - \cos^2 = 0

    We set u = cos^2 and get u^2 - u = 0 which gives us u = 0 and u = 1.

    cos^2 x = 0 gives us cos x = +/- 0. x = pi/2 + k(pi).
    cos^2 x = 1 gives us cos x = +/- 1. x = 0 + k(pi).

    So the solutions are, in order,
    0, pi/2, pi, 3pi/2.

    I apologize for the mix and match between Latex and regular text.
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  2. #2
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    Quote Originally Posted by Sabo View Post
    I did get the right answers, but I still would like to make sure my thinking is accurate, and also check if I made unnecessary calculations somewhere.

    Get all the solutions to the trigonometric equation \sin^4 x + \cos^4 x = 1 in the interval [0, 2pi[

    What I did was:

    \sin x\sin x\sin x\sin x + \cos x\cos x\cos x\cos x = 1

    <br />
[-\dfrac{1}{2}(\cos 2x - \cos 0)][-\dfrac{1}{2}(\cos 2x - \cos 0)] + [\dfrac{1}{2}(\cos 2x + \cos 0)][\dfrac{1}{2}(\cos 2x + \cos 0)] = 1

    <br />
[-\dfrac{1}{2}(2\cos^2 x - 2)][-\dfrac{1}{2}(2\cos^2 x - 2)] + [\dfrac{1}{2}(2\cos^2 x)][\dfrac{1}{2}(2\cos^2 x)] = 1

    (-\cos^2 x - 1)^2 + \cos^4 x = 1

    2\cos^4 x - 2\cos^2 x + 1 = 1

    \cos^4 x - \cos^2 = 0

    We set u = cos^2 and get u^2 - u = 0 which gives us u = 0 and u = 1.

    cos^2 x = 0 gives us cos x = +/- 0. x = pi/2 + k(pi).
    cos^2 x = 1 gives us cos x = +/- 1. x = 0 + k(pi).

    So the solutions are, in order,
    0, pi/2, pi, 3pi/2.

    I apologize for the mix and match between Latex and regular text.
    This is easier if you make use of the Pythagorean Identity...


    \sin^4{x} + \cos^4{x} = 1

    (\sin^2{x})^2 + (\cos^2{x})^2 = 1

    (\sin^2{x})^2 + (1 - \sin^2{x})^2 = 1

    (\sin^2{x})^2 + 1 - 2\sin^2{x} + (\sin^2{x})^2 = 1

    2(\sin^2{x})^2 - 2\sin^2{x} = 0

    2\sin^2{x}(\sin^2{x} - 1) = 0.


    Case 1: \sin^2{x} = 0

    \sin{x} = 0

    x = \pi n.


    Case 2: \sin^2{x} - 1 = 0

    \sin^2{x} = 1

    \sin{x} = \pm 1

    x = \frac{\pi}{2} + \pi n.


    So the solutions are:

    x = \left\{0, \frac{\pi}{2}\right\} + \pi n

    x = \left\{ 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \right\}

    (I can't read if you are allowing 2\pi to be in the interval... If it's not then get rid of it).
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  3. #3
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    Ah, right. Thanks.
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