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Thread: 3 Trigonometric Equations

  1. #1
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    2 Trigonometric Equations

    I have next week an exam and I don't know how to solve the following 2 equations. If anyone knows how to solve them, please, help me.

    sin2(4x) sin2(2x) = 0
    cos(2x) = cos(x) sin(x)

    The teacher gave me the solutions but I don't know how to get them. Here are the solutions:
    Last edited by faris92; Nov 26th 2009 at 10:11 AM.
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  2. #2
    Senior Member I-Think's Avatar
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    I'll handle the 1st question,
    $\displaystyle sin^2(4x)-sin^2(2x)=0$

    $\displaystyle (2sin2xcos2x)^2=(2sinxcosx)^2$

    $\displaystyle sin2xcos2x=sinxcosx$

    $\displaystyle 2sinxcosx(cos^2x-sin^2x)=sinxcosx$

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

    $\displaystyle 2cos^2x-1=\frac{1}{2}$

    $\displaystyle cos^2x=\frac{3}{4}$

    $\displaystyle cosx=\pm{\frac{\sqrt{3}}{2}}$

    Values of x that satisfy this equation are $\displaystyle x=\frac{\pi}{6}$ and $\displaystyle x=\frac{5\pi}{6}$. And as the $\displaystyle cos(x)$ function is periodic, as a consequence all integer multiples of those functions are your solution.
    And that's how you get your answer for number 1.
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  3. #3
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    Here's a solution to the second one.

    I've doublechecked this and am pretty sure I haven't made any errors.

    However, the answers your teacher gave you don't work when I plug them in.

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

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

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

    $\displaystyle 1-4\sin^2{x}+4\sin^4{x}=1-2\cos{x}\sin{x}$

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

    $\displaystyle 4\sin^2{x}(\sin^2{x}-1)=-2\cos{x}\sin{x}$

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

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

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

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

    $\displaystyle x=\frac{\pi}{4}$ and $\displaystyle x=\frac{\pi}{4}+\pi$ satisfy this equation.
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