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Math Help - Solving Trigonometric equation

  1. #1
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    Solving Trigonometric equation

    Hi,
    Looking for help with this problem---I'm stumped because I get different answers than the book.
    Here's the problem:
    (2sin^2)(2x) = 1

    The book gets the following answers: pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8,11pi/8, 13pi/8, 15pi/8.

    I keep getting pi/4, 3pi/4, 5pi/4, 7pi/4.

    Appreciate an explanation of how they get the answers.
    Mike Clemmons
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, Mike!

    You're probably making a small error.

    I assume that answers are in the interval [0,\:2\pi]

    I'll take baby steps . . .


    2\sin^2(2x) \:=\: 1

    Divide by 2: . \sin^2(2x) \:=\:\frac{1}{2}

    Take square roots: . \sin(2x) \;=\;\pm\frac{1}{\sqrt{2}}

    (Can you see that the angle is a 45 angle in each quadrant?)

    We have: . 2x \;=\;\frac{\pi}{4} + \frac{\pi}{2}n

    Divide by 2: . x \;=\;\frac{\pi}{8} + \frac{\pi}{4}n


    Let n \:=\:0,1,2,3,\hdots 7

    . . x \;=\;\frac{\pi}{8},\;\frac{3\pi}{8},\;\frac{5\pi}{  8},\;\frac{7\pi}{8},\;\frac{9\pi}{8},\;\frac{11\pi  }{8},\;\frac{13\pi}{8},\;\frac{15\pi}{8}

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