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Thread: trig equation

  1. #1
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    trig equation

    Hi I am not sure what substitution to use for this, have tried $\displaystyle cos(x+x)=cosxcosx-sinxsinx$ and then I tried $\displaystyle cos\theta+cosy=2cos(\frac{x+y}{2})cos(\frac{x-y}{2})$

    the question is: find all solutions if $\displaystyle 2cos2x+4sinxcosx=\sqrt{2}$ in the range $\displaystyle o \leq x \leq 2\pi$
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by oxrigby View Post
    Hi I am not sure what substitution to use for this, have tried $\displaystyle cos(\theta+\theta)=cos\thetacos\theta-sin\thetasin\theta$ and then I tried $\displaystyle cos\theta+cosy=2cos(\frac{x+y}{2})cos(\frac{x-y}{2})$

    the question is: find all solutions if $\displaystyle 2cos2x+4sinxcosx=\sqrt{2}$ in the range $\displaystyle \leg\theta\leq2\pi$
    $\displaystyle 2\cos(2x)+4\sin(x)\cos(x)=\sqrt{2}\Rightarrow{\cos (2x)+\sin(2x)=\frac{\sqrt{2}}{2}}$

    Does that help?
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  3. #3
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    My original post is latex sorted now,,,
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  4. #4
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    ok
    Last edited by oxrigby; Nov 10th 2008 at 04:05 PM.
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