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Math Help - [SOLVED] Trig equation

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Trig equation

    tan(2x) = \frac{1 + sin(x)}{cos(x)}

    I'm all out of ideas here. Anyone?
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  2. #2
    MHF Contributor
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    I'd start with the Right-hand side. Multiply numerator and denominator by 1 + sin(x) and see what happens.

    Like many such problems, it may not lead anywhere, but it's a place to start and something else may hit you.

    For example, while I was typing this, it dawned on me that I don't happen to know any formulas for tan(2x) right off the top of my head. I might be tempted to change that to sines and cosines. Maybe not.

    Rule #1 on Trig Identities -- You can't break it. Just try something.
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  3. #3
    Senior Member Spec's Avatar
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    Thanks, but I've already tried everything I can think of. I need to see a solution soon or my head will explode.

    As if the last one wasn't bad enough, I'm now stuck with this one as well:

    2cos^3(x) + sin(x) = 2cos(x)
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  4. #4
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    Hello, Spec!

    2\cos^3\!x + \sin x \:= \:2\cos x

    We have: . 2\cos x(\cos^2\!x) + \sin x \;=\;2\cos x \quad\Rightarrow\quad 2\cos x(1 - \sin^2\!x) + \sin x \;=\;2\cos x

    . . . . 2\cos x - 2\sin^2\!x\cos x + \sin x \;=\;2\cos x \quad\Rightarrow\quad \sin x - 2\sin^2\!x\cos x \;=\;0

    Factor: . \sin x(1 - 2\sin x\cos x) \;=\;0


    And we have two equations to solve:

    \sin x \:=\:0\quad\Rightarrow\quad\boxed{x \:=\:\pi n}

    1 - 2\sin x\cos x\:=\:0\quad\Rightarrow\quad 1 - \sin2x \:=\:0\quad\Rightarrow\quad \sin2x \:=\:1
    . . 2x \:=\:\frac{\pi}{2} + 2\pi n \quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{4} + \pi n}

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  5. #5
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    Hello, Spec!

    I'll give only the principal roots.
    . . You can generalize the answers . . .


    Solve for x\!:\;\;\tan 2x \;= \;\frac{1 + \sin x}{\cos x}

    We have: . \frac{\sin2x}{\cos2x} \;=\;\frac{1+\sin x}{\cos x} \quad\Rightarrow\quad \frac{2\sin x\cos x}{1 - 2\sin^2\!x} \;=\;\frac{1+\sin x}{\cos x}

    . . 2\sin x\cos^2x \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x

    . . 2\sin x(1 - \sin^2\!x) \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x

    . . 2\sin x - 2\sin^3\!x \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x

    . . which simplifies to: . 2\sin^2\!x + \sin x - 1 \;=\;0

    . . and factors: . (2\sin x - 1)(\sin x + 1) \;=\;0


    And we have two equations to solve.

    . . 2\sin x - 1 \:=\:0\quad\Rightarrow\quad \sin x \:=\:\frac{1}{2}\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{6},\:\frac{5\pi}{6}}

    . . \sin x + 1\:=\:0\quad\Rightarrow\quad \sin x \:=\:-1\quad\Rightarrow\quad\boxed{ x \:=\:\frac{3\pi}{2}}

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  6. #6
    Senior Member Spec's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Spec!

    I'll give only the principal roots.
    . . You can generalize the answers . . .



    We have: . \frac{\sin2x}{\cos2x} \;=\;\frac{1+\sin x}{\cos x} \quad\Rightarrow\quad \frac{2\sin x\cos x}{1 - 2\sin^2\!x} \;=\;\frac{1+\sin x}{\cos x}

    . . 2\sin x\cos^2x \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x

    . . 2\sin x(1 - \sin^2\!x) \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x

    . . 2\sin x - 2\sin^3\!x \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x

    . . which simplifies to: . 2\sin^2\!x + \sin x - 1 \;=\;0

    . . and factors: . (2\sin x - 1)(\sin x + 1) \;=\;0


    And we have two equations to solve.

    . . 2\sin x - 1 \:=\:0\quad\Rightarrow\quad \sin x \:=\:\frac{1}{2}\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{6},\:\frac{5\pi}{6}}

    . . \sin x + 1\:=\:0\quad\Rightarrow\quad \sin x \:=\:-1\quad\Rightarrow\quad\boxed{ x \:=\:\frac{3\pi}{2}}

    Thanks! x = \frac{3\pi}{2} + 2{\pi}n is not a solution though, since cos x \neq 0.
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