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Math Help - Solve this trig equation

  1. #1
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    Solve this trig equation

    I completed all of the harder questions...but for some reason I'm stuck on the first problem in my assignment.

    Can someone help?

     sin^2\theta = 2cos^2\theta

    haha thanks
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  2. #2
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    \displaystyle \sin^2{\theta} = 2\cos^2{\theta}

    \displaystyle 1 - \cos^2{\theta} = 2\cos^2{\theta}

    \displaystyle 1 = 3\cos^2{\theta}

    \displaystyle \frac{1}{3} = \cos^2{\theta}

    \displaystyle \pm \frac{1}{\sqrt{3}} = \cos{\theta}

    \displaystyle \theta = \arccos{\left(\pm\frac{1}{\sqrt{3}}\right)}


    So the solutions are: \displaystyle \theta = \left\{ \arccos{\left(\frac{1}{\sqrt{3}}\right)}, \pi - \arccos{\left(\frac{1}{\sqrt{3}}\right)}, \pi + \arccos{\left(\frac{1}{\sqrt{3}}\right)}, 2\pi - \arccos{\left(\frac{1}{\sqrt{3}}\right)} \right\} + 2\pi n where \displaystyle n \in \mathbf{Z}.
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  3. #3
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    Ah, I forgot about sin^2\theta+cos^2\theta=1 haha

    Thank you!
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  4. #4
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    Hello, jellyksong!

     \sin^2\!\theta \:=\: 2\cos^2\!\theta

    \text{Divide by }\cos^2\!\theta:\;\;\dfrac{\sin^2\!\theta}{\cos^2\  !\theta} \,=\,2\quad\Rightarrow\quad \left(\dfrac{\sin\theta}{\cos\theta}\right)^2 =\,2 \quad\Rightarrow\quad \tan^2\!\theta \,=\,2


    . . . . \tan\theta \,=\,\pm\sqrt{2} \quad\Rightarrow\quad \theta \:=\:\tan^{-1}\!\left(\pm\sqrt{2}\right)

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, jellyksong!


    \text{Divide by }\cos^2\!\theta:\;\;\dfrac{\sin^2\!\theta}{\cos^2\  !\theta} \,=\,2\quad\Rightarrow\quad \left(\dfrac{\sin\theta}{\cos\theta}\right)^2 =\,2 \quad\Rightarrow\quad \tan^2\!\theta \,=\,2


    . . . . \tan\theta \,=\,\pm\sqrt{2} \quad\Rightarrow\quad \theta \:=\:\tan^{-1}\!\left(\pm\sqrt{2}\right)

    Thou shalt not divide by 0...
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Thou shalt not divide by 0...
    Amen.
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  7. #7
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    Hello, Prove it!

    Quote Originally Posted by Prove It
    Thou shalt not divide by 0...

    Good point . . .

    Since neither \dfrac{\pi}{2} nor \dfrac{3\pi}{2} are possible roots, I thought I was safe.

    But I should have said so.

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