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Thread: Trig equations: cos4x=sin2x

  1. #1
    RAz
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    Trig equations: cos4x=sin2x

    I have to solve cos4x=sin2x where x is an element of 0, 360.

    The problem, for me, is cos4x. I haven't proved this myself, but a quick search gave me that cos4x=$\displaystyle 8cos^4x-8cos^2x+1$

    By going along the lines of $\displaystyle 8cos^4x-8cos^2x+1=2sinxcosx$ (see original question) I managed to arrive at sinx=$\displaystyle \frac{1}{3}$

    This does not give an exact value and I was wondering if I had done this correctly. Thanks!
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  2. #2
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    Quote Originally Posted by RAz View Post
    I have to solve cos4x=sin2x where x is an element of 0, 360.

    The problem, for me, is cos4x. I haven't proved this myself, but a quick search gave me that cos4x=$\displaystyle 8cos^4x-8cos^2x+1$

    By going along the lines of $\displaystyle 8cos^4x-8cos^2x+1=2sinxcosx$ (see original question) I managed to arrive at sinx=$\displaystyle \frac{1}{3}$

    This does not give an exact value and I was wondering if I had done this correctly. Thanks!
    $\displaystyle \cos{4x} = \sin{2x}$

    $\displaystyle \cos^2{2x} - \sin^2{2x} = \sin{2x}$

    $\displaystyle 1 - \sin^2{2x} - \sin^2{2x} = \sin{2x}$

    $\displaystyle 1 - 2\sin^2{2x} = \sin{2x}$

    $\displaystyle 0 = 2\sin^2{2x} + \sin{2x} - 1$.


    You now have a quadratic equation in $\displaystyle \sin{2x}$.

    So let $\displaystyle \sin{2x} = X$, so that you can see the Quadratic:


    $\displaystyle 0 = 2X^2 + X - 1$

    $\displaystyle 0 = 2X^2 + 2X - X - 1$

    $\displaystyle 0 = 2X(X + 1) - 1(X + 1)$

    $\displaystyle 0 = (X + 1)(2X - 1)$.


    So $\displaystyle X = -1$ or $\displaystyle X = \frac{1}{2}$.


    This means $\displaystyle \sin{2x} = -1$ or $\displaystyle \sin{2x} = \frac{1}{2}$.


    Case 1:

    $\displaystyle \sin{2x} = -1$

    $\displaystyle 2x = 270^{\circ} + 360^{\circ}n$, where $\displaystyle n$ is an integer.

    $\displaystyle x = 135^{\circ} + 180^{\circ}n$, where $\displaystyle n$ is an integer.


    Case 2:

    $\displaystyle \sin{2x} = \frac{1}{2}$

    Since sine is positive in the first and second quadrants:

    $\displaystyle 2x = \left\{30^{\circ}, 180^{\circ} - 30^{\circ} \right\} + 360^{\circ} n$

    $\displaystyle 2x = \left\{ 30^{\circ}, 150^{\circ}\right\} + 360^{\circ} n$

    $\displaystyle x = \left \{ 15^{\circ}, 75^{\circ} \right\} + 180^{\circ} n$.



    So putting it together:

    $\displaystyle x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}\right\} + 180^{\circ}n$.


    Upon substituting $\displaystyle n = 0$ and $\displaystyle n = 1$, we find all the possible solutions in the domain $\displaystyle 0^{\circ} \leq x \leq 360^{\circ}$:


    $\displaystyle x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}, 195^{\circ}, 255^{\circ}, 315^{\circ} \right\}$.
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  3. #3
    RAz
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    Thanks a lot, I appreciate it. Very clear.
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    Just to give an alternative way

    $\displaystyle \cos{4x} = \sin{2x}$

    $\displaystyle \cos{4x} = \cos{(90^{\circ}-2x)}$

    Case 1:
    $\displaystyle 4x = 90^{\circ}-2x+ 360^{\circ} n$

    $\displaystyle 6x = 90^{\circ} + 360^{\circ} n$

    $\displaystyle x = 15^{\circ} + 60^{\circ} n$

    $\displaystyle x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}, 195^{\circ}, 255^{\circ}, 315^{\circ} \right\}$


    Case 2:
    $\displaystyle 4x = -(90^{\circ}-2x) + 360^{\circ} n$

    $\displaystyle 4x = -90^{\circ}+2x + 360^{\circ} n$

    $\displaystyle 2x = -90^{\circ} + 360^{\circ} n$

    $\displaystyle x = -45^{\circ} + 180^{\circ} n$

    $\displaystyle x = \left\{135^{\circ}, 315^{\circ} \right\}$


    Therefore:
    $\displaystyle x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}, 195^{\circ}, 255^{\circ}, 315^{\circ} \right\}$
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