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Thread: Solve for theta

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    Solve for theta

    Write all positive values within the interval of 0 degrees < $\displaystyle \theta$ < 270 degrees

    $\displaystyle 1 - sec^{2} 2\theta - 3tan 2\theta = 0$
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  2. #2
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    Quote Originally Posted by reiward View Post
    Write all positive values within the interval of 0 degrees < $\displaystyle \theta$ < 270 degrees

    $\displaystyle 1 - sec^{2} 2\theta - 3tan 2\theta = 0$
    Substitute $\displaystyle \sec^2 (2 \theta) = \tan^2 (2 \theta) + 1$, simplify and solve the resulting simple trig equation.
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    Quote Originally Posted by mr fantastic View Post
    Substitute $\displaystyle \sec^2 (2 \theta) = \tan^2 (2 \theta) + 1$, simplify and solve the resulting simple trig equation.
    $\displaystyle
    1 - sec^{2} 2\theta - 3tan 2\theta = 0$

    $\displaystyle
    1 - (\tan^2 (2 \theta) + 1) - 3tan 2\theta = 0$

    $\displaystyle \tan^2 (2 \theta) - 3tan 2\theta = 0$

    Im stuck. What do I factor it to?
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    Quote Originally Posted by reiward View Post
    $\displaystyle
    1 - sec^{2} 2\theta - 3tan 2\theta = 0$

    $\displaystyle
    1 - (\tan^2 (2 \theta) + 1) - 3tan 2\theta = 0$

    $\displaystyle \tan^2 (2 \theta) - 3tan 2\theta = 0$ Mr F says: This is wrong. There is a careless mistake in signs.

    Im stuck. What do I factor it to?
    Factorise. Note that $\displaystyle \tan (2 \theta)$ is a common factor. Then use the Null Factor Law to get two simple trig equations. Solve them.
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    Quote Originally Posted by mr fantastic View Post
    Substitute $\displaystyle \sec^2 (2 \theta) = \tan^2 (2 \theta) + 1$, simplify and solve the resulting simple trig equation.
    Quote Originally Posted by mr fantastic View Post
    Factorise. Note that $\displaystyle \tan (2 \theta)$ is a common factor. Then use the Null Factor Law to get two simple trig equations. Solve them.
    Oh okay.

    $\displaystyle -\tan^2 (2 \theta) - 3tan 2\theta = 0$

    Can I make that to $\displaystyle \tan^2 (2 \theta) + 3tan 2\theta = 0$ ?

    If so, would the factor be $\displaystyle \tan(2 \theta) [tan + 3]= 0$?
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    Quote Originally Posted by reiward View Post
    Oh okay.

    $\displaystyle -\tan^2 (2 \theta) - 3tan 2\theta = 0$

    Can I make that to $\displaystyle \tan^2 (2 \theta) + 3tan 2\theta = 0$ ?

    If so, would the factor be $\displaystyle \tan(2 \theta) [tan + 3]= 0$?
    Yes.
    Yes.
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