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Thread: Please help.Trignometric equations.

  1. #1
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    Please help.Trignometric equations.

    Can someone help me on these

    1.$\displaystyle cos2x=sin3x$

    2.$\displaystyle \sqrt{3}tanx-secx-1=0$

    3.$\displaystyle sec3\theta=sec\theta$

    4.$\displaystyle sin4x-sin2x=cos3x$


    thanks
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: Please help.Trignometric equations.

    first one 2x+3x = 90, or pi/2. and same angle in the third quarter with the 2n pi

    the seond one use the identity

    tan^2 x + 1 = sec ^2 x
    Last edited by Amer; Aug 15th 2012 at 04:37 PM.
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  3. #3
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    Re: Please help.Trignometric equations.

    Quote Originally Posted by hacker804 View Post
    Can someone help me on these

    1.$\displaystyle cos2x=sin3x$

    2.$\displaystyle \sqrt{3}tanx-secx-1=0$

    3.$\displaystyle sec3\theta=sec\theta$

    4.$\displaystyle sin4x-sin2x=cos3x$


    thanks
    1.
    $\displaystyle \displaystyle \begin{align*} \cos{2x} &= \sin{3x} \\ 1 - 2\sin^2{x} &= 3\sin{x} - 4\sin^3{x} \\ 4\sin^3{x} - 2\sin^2{x} - 3\sin{x} + 1 &= 0 \end{align*}$

    Write $\displaystyle \displaystyle \begin{align*} X = \sin{x} \end{align*}$ to get the polynomial $\displaystyle \displaystyle \begin{align*} 4X^3 - 2X^2 - 3X + 1 = 0 \end{align*}$.

    It's easy to see that $\displaystyle \displaystyle \begin{align*} X = 1 \end{align*}$ satisfies this equation, so $\displaystyle \displaystyle \begin{align*} X - 1 \end{align*}$ is a factor, and long division gives

    $\displaystyle \displaystyle \begin{align*} 4X^3 - 2X^2 - 3X + 1 &= 0 \\ (X - 1)\left(4X^2 + 2X - 1\right) &= 0 \\ 4(X - 1)\left(X^2 + \frac{1}{2}X - \frac{1}{4}\right) &= 0 \\ 4(X - 1)\left[X^2 + \frac{1}{2}X + \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2 - \frac{1}{4}\right] &= 0 \\ 4(X - 1)\left[\left(X + \frac{1}{4}\right)^2 - \frac{5}{16} \right] &= 0 \\ 4(X - 1)\left[ \left( X + \frac{1}{4} \right)^2 - \left( \frac{\sqrt{5}}{4} \right)^2 \right] &= 0 \\ 4(X - 1)\left( X + \frac{1}{4} - \frac{\sqrt{5}}{4} \right)\left( X + \frac{1}{4} + \frac{\sqrt{5}}{4} \right) &= 0 \\ X - 1 = 0 \textrm{ or } X + \frac{1}{4} - \frac{\sqrt{5}}{4} \textrm{ or } X + \frac{1}{4} + \frac{\sqrt{5}}{4} &= 0 \\ X = 1 \textrm{ or } X = \frac{-1 + \sqrt{5}}{4} \textrm{ or } X &= \frac{-1 - \sqrt{5}}{4} \end{align*}$

    So you need to solve $\displaystyle \displaystyle \begin{align*} \sin{x} = 1 \end{align*}$ (easy), as well as $\displaystyle \displaystyle \begin{align*} \sin{x} = \frac{1 + \sqrt{5}}{4} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \sin{x} = \frac{1 - \sqrt{5}}{4} \end{align*}$ (not so easy). All will have exact value answers though. This will help you.
    Last edited by Prove It; Aug 15th 2012 at 05:07 PM.
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  4. #4
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    Re: Please help.Trignometric equations.

    √3 tan⁡x-sec⁡x-1=0
    √3 sin⁡x/cos⁡x -1/cos⁡x -1=0
    √3 sin⁡x-1- cos⁡x=0
    √3/2 sin⁡x-1/2 cos⁡x=1/2 {2 is √((√3)^2+(-1)^2 )}
    cos⁡〖π/6 sin⁡x-〗 sin⁡〖π/6〗 cos⁡x=sin⁡〖π/6〗
    sin⁡〖(π/6-x)=〗 sin⁡〖π/6〗
    π/6-x=nπ+(-1)^n π/6
    x=π/6-nπ-(-1)^n π/6
    here n is a integer
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  5. #5
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    Re: Please help.Trignometric equations.

    sec⁡θ=sec⁡3θ
    cos⁡θ=cos⁡3θ
    cos⁡θ=4cos^3⁡θ-3 cos⁡θ
    0=4 cos^3⁡θ-4 cos⁡θ
    4 cos⁡θ {cos^2⁡θ-1}=0
    cos⁡θ (cos⁡θ-1)(cos⁡θ+1)=0
    cos⁡θ=0 or cos⁡θ=1 or cos⁡θ=-1
    cos⁡θ=cos⁡0 or cos⁡θ=cos⁡〖π/2〗 or cos⁡θ=cos⁡((-π)/2)
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