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Thread: Rewrite this identity as a quadratic equation in tan(x)

  1. #1
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    Rewrite this identity as a quadratic equation in tan(x)

    "Show that the equation $\displaystyle sin^2x + 3 sin\ x\ cos\ x = 2$ can be written as a quadratic equation in $\displaystyle tan\ x$."

    I can't see any way to get a quadratic equation with no other trig functions than $\displaystyle tan\ x$. I tried dividing throughout by $\displaystyle cos\ x$ etc. to generate $\displaystyle tan\ x$, but always ended up with a leftover sin or cos somewhere.

    Could anyone give me some hints?
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by EvilKitty View Post
    "Show that the equation $\displaystyle sin^2x + 3 sin\ x\ cos\ x = 2$ can be written as a quadratic equation in $\displaystyle tan\ x$."

    I can't see any way to get a quadratic equation with no other trig functions than $\displaystyle tan\ x$. I tried dividing throughout by $\displaystyle cos\ x$ etc. to generate $\displaystyle tan\ x$, but always ended up with a leftover sin or cos somewhere.

    Could anyone give me some hints?
    Divide the whole equation by $\displaystyle \cos^2 x$, and remember that $\displaystyle \frac{1}{\cos^2 x}=1+\tan^2 x$.
    Note that you don't lose anything by doing that division: an x for which $\displaystyle \cos x=0$ could not possibly be a solution of the original equation.
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  3. #3
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    It worked, thanks! I got $\displaystyle -tan^2\ x\ + 3tan\ x -2 = 0$.

    How did you get $\displaystyle \frac{1}{\cos^2 x}=1+\tan^2 x$? I can't see how to derive this from the trigonometric identities I've learnt so far.
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by EvilKitty View Post
    It worked, thanks! I got $\displaystyle -tan^2\ x\ + 3tan\ x -2 = 0$.

    How did you get $\displaystyle \frac{1}{\cos^2 x}=1+\tan^2 x$? I can't see how to derive this from the trigonometric identities I've learnt so far.
    Just use the fact that $\displaystyle \cos^2x+\sin^2x = 1$, for all x, to replace the numerator by something a little more interesting:

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