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Thread: Solve a trigonomoetric equation

  1. November 13th 2008, 07:57 AM #1
    Cam5
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    Solve a trigonomoetric equation

    Hi there - I need to Solve 2sin (2θ) = 1 for 0<x<2pi

    I've tried this so far:

    2sin2x = 1
    sin2x = 1/2
    2sinxcosx = 1/2
    sinxcosx=1/4
    sinxcosx - 1/4 = 0

    But I don't know how to factor this, if I should be factoring, or how to get all the possible solutions for x... Could someone give me a hand?

    - Cam
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  2. November 13th 2008, 09:13 AM #2
    masters
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    Quote Originally Posted by Cam5 View Post
    Hi there - I need to Solve 2sin (2θ) = 1 for 0<x<2pi

    I've tried this so far:

    2sin2x = 1
    sin2x = 1/2
    2sinxcosx = 1/2
    sinxcosx=1/4
    sinxcosx - 1/4 = 0

    But I don't know how to factor this, if I should be factoring, or how to get all the possible solutions for x... Could someone give me a hand?

    - Cam
    2\sin(2\theta)=1

    \sin(2\theta)=\frac{1}{2}

    If we restrict the domain of the sine function to:

    \left[-\frac{\pi}{2}\leq 2x \leq \frac{\pi}{2}\right]\rightarrow \left[-\frac{\pi}{4}\leq x \leq \frac{\pi}{4}\right],

    we can use the inverse sine function to solve for reference angle 2x and then x.

    \sin^{-1}(\sin(2x))=\sin^{-1}\left(\frac{1}{2}\right)

    2x=\sin^{-1}\left(\frac{1}{2}\right)

    x=\frac{1}{2} \sin^{-1} \left(\frac{1}{2}\right)

    x\approx .2617993878 radians.
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  3. November 13th 2008, 09:24 AM #3
    Soroban
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    Hello, Cam5!

    You started off fine . . .

    Solve: . 2\sin(2\theta) \:=\:1 .for 0\,<\,x\,<\,2\pi

    I've tried this so far: . 2\sin(2\theta) \:=\: 1 \quad\Rightarrow\quad \sin(2\theta) \:=\: \tfrac{1}{2}

    I'm delighted that you know that double-angle identity,
    . . but that's not the way to go . . .


    We have: . \sin(2\theta) \:=\:\tfrac{1}{2}

    . . .Then: . 2\theta \:=\:\frac{\pi}{6},\:\frac{5\pi}{6},\:\frac{13\pi} {6},\:\frac{17\pi}{6},\:\frac{25\pi}{6},\:\frac{29 \pi}{6},\:\hdots

    . . . \text{Hence: }\;\theta \;=\;\underbrace{\frac{\pi}{12},\:\frac{5\pi}{12}, \:\frac{13\pi}{12},\:\frac{17\pi}{12}}_{\text{Thes e are in the interval}},\:\frac{25\pi}{12},\:\frac{29\pi}{12},\ :\hdots


    Therefore: . \theta \;=\;\frac{\pi}{12},\;\frac{5\pi}{12},\;\frac{13\p i}{12},\;\frac{17\pi}{12}

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