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Thread: solving the equation

  1. #1
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    solving the equation

    Hi there,

    it would be much appreciated if any one can help me solve this equation

    2tan^2x+2sec^2x-secx =12

    thanks
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \frac{2\sin^2x}{\cos^2x}+\frac{2}{\cos^2x}-\frac{1}{\cos x}=12\Rightarrow 12\cos^2x+\cos x-2\sin^2x-2=0$

    Replace $\displaystyle \sin^2x=1-\cos^2x$, then note $\displaystyle \cos x=t$ and solve the quadratic.

    Can you do that?
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  3. #3
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    Quote Originally Posted by red_dog View Post
    $\displaystyle \frac{2\sin^2x}{\cos^2x}+\frac{2}{\cos^2x}-\frac{1}{\cos x}=12\Rightarrow 12\cos^2x+\cos x-2\sin^2x-2=0$

    Replace $\displaystyle \sin^2x=1-\cos^2x$, then note $\displaystyle \cos x=t$ and solve the quadratic.

    Can you do that?
    sorry but i dont know wat u mean when u say ''note cos x =t
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  4. #4
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    Quote Originally Posted by masiboy View Post
    sorry but i dont know wat u mean when u say ''note cos x =t
    red_dog means make the substitution $\displaystyle t = \cos x$ and solve the resulting quadratic equation for t. Then, once you have found the value(s) of t, substitute back into $\displaystyle t = \cos x$ and solve for x.
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  5. #5
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    Hello, masiboy!

    I'll assume that the answers are on $\displaystyle [0^o,\,360^o]$


    $\displaystyle 2\tan^2\!x+2\sec^2\!x-\sec x \:=\:12$

    Since $\displaystyle \tan^2\!x \:=\:\sec^2\!x - 1$, the equation becomes: .$\displaystyle 2(\sec^2\!x - 1) + 2\sec^2\!x - \sec x \:=\:12$

    . . which simplifies to: .$\displaystyle 4\sec^2\!x - \sec x - 14 \:=\:0$

    . . and factors: .$\displaystyle (\sec x - 2)(4\sec x + 7) \:=\:0$


    We have: .$\displaystyle \sec x -2 \:=\:0 \quad\Rightarrow\quad \sec x \:=\:2\quad\Rightarrow\quad x \:=\:60^o,\:300^o$

    . . and: .$\displaystyle 4\sec x + 7 \:=\:\quad\Rightarrow\quad \sec x \:=\:-\frac{7}{4}\quad\Rightarrow\quad x \;\approx\;124.85^o,\:235.15^o$

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