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Math Help - (tan x)^2 = sin 2x

  1. #1
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    (tan x)^2 = sin 2x

    anyone has an idea on how to solve for this equation, (tan x)^2 = sin 2x

    I know that sin 2x = 2sin xcos x and tan x = sin x/cos x, how ever i am having trouble simplifyng the equation into a single function so that I can solve for x.

    thanks in advance

    @sammys thanks for your advice on the other thread.
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  2. #2
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    Re: (tan x)^2 = sin 2x

    \displaystyle \begin{align*} \tan^2{x} &= \sin{2x} \\ \frac{\sin^2{x}}{\cos^2{x}} &= 2\sin{x}\cos{x} \\ \sin^2{x} &= 2\sin{x}\cos^3{x} \\ \sin^2{x} - 2\sin{x}\cos^3{x} &= 0 \\ \sin{x}\left(\sin{x} - 2\cos^3{x}\right) &= 0 \\ \sin{x} = 0 \textrm{ or }\sin{x} - 2\cos^3{x} &= 0 \end{align*}

    Can you solve each of those two equations now?
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  3. #3
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    Re: (tan x)^2 = sin 2x

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} \tan^2{x} &= \sin{2x} \\ \frac{\sin^2{x}}{\cos^2{x}} &= 2\sin{x}\cos{x} \\ \sin^2{x} &= 2\sin{x}\cos^3{x} \\ \sin^2{x} - 2\sin{x}\cos^3{x} &= 0 \\ \sin{x}\left(\sin{x} - 2\cos^3{x}\right) &= 0 \\ \sin{x} = 0 \textrm{ or }\sin{x} - 2\cos^3{x} &= 0 \end{align*}

    Can you solve each of those two equations now?
    solving for x:

    arcsin(0) = x

    x = 0 + \pi k, ~ k\in \mathbb{Z}

    solving for sin(x):

    can you say that " sin(x) = sin(\pi k), ~ k\in \mathbb{Z}"?

    How would you go about solving the 2nd equation? Sorry OP if I'm hijacking; I am interested in your question too
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: (tan x)^2 = sin 2x

    For the first one you can just write for the solution: x=k\pi with k\in \mathbb{Z}\right
    I've no clue to solve the second one, I tried something but it doesn't seem to work. Probably another member has a good solution for it .
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  5. #5
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    Re: (tan x)^2 = sin 2x

    @prove it..

    I arrived at the fourth line of the solution you had presented, and i never thought of factoring out sin x, i missed that part. yes, I can solve the remaining equations now. a million thanks to you.
    to otherw whoe replied, thank you.
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  6. #6
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    Re: (tan x)^2 = sin 2x

    what i did actually before is that i cancelled the sin x part on the right hand side that is why i am having difficulty finding out the solution to the equation. thank you guys
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  7. #7
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    Re: (tan x)^2 = sin 2x

    oops.. the second equation, i realized is the equation I am stuck with.. sin x - 2(cos x)^3 = 0

    sorry.. still thinking of the solution
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  8. #8
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    Re: (tan x)^2 = sin 2x

    Quote Originally Posted by bigjoe5263 View Post
    oops.. the second equation, i realized is the equation I am stuck with.. sin x - 2(cos x)^3 = 0

    sorry.. still thinking of the solution
    Divide by cos(x), which is fine if cos(x) ≠ 0. And clearly, cos(x) = 0 is not a solution to this equation.

    sin(x)/cos(x) = tan(x).

    Multiply by sec^2(x), and change sec^2(x) to 1 + tan^2(x).

    You then have a cubic equation in tan(x). Let u = tan(x), & factor to solve for u.
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  9. #9
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    Re: (tan x)^2 = sin 2x

    Quote Originally Posted by SammyS View Post
    Divide by cos(x), which is fine if cos(x) ≠ 0. And clearly, cos(x) = 0 is not a solution to this equation.
    Just wondering why it is clear that cos(x) = 0 could not be a solution?
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: (tan x)^2 = sin 2x

    If you solve \cos(x)=0 \Leftrightarrow x=\frac{\pi}{2}+2k\pi
    And \tan\left(\frac{\pi}{2}+2k\pi\right) is not defined.
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