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Thread: solving trig equations

  1. #1
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    solving trig equations

    can somebody please help me
    I have the question and answer to this question but i do not understand at all how the answer is concluded

    Q: in the range 0<=x<=2Pi solve Sin2x=-1
    A:3Pi/8 and 7Pi/8

    could somebody maybe briefly explain how you work this out

    thanks
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  2. #2
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    For angles ranging from 0 to 360,
    Sin(angle) may be viewed as the y co-ordinate of a point on a unit circle,
    of radius 1.

    If 2x is the angle,
    where is the y co-ordinate -1 ?

    answer: at 270 degrees, or $\displaystyle \frac{3}{4}2\pi$

    Sin(270 degrees)= -1.

    So 2x is 270 degrees or 270+n(360) degrees, n=0,1,2,3........

    x is $\displaystyle \frac{3}{4}\pi$

    The second result is obtained for 2x=270+360 degrees,
    as x<360 degrees.
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  3. #3
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    Hello, wahhdoe!

    Solve: .$\displaystyle \sin2x \:=\:\text{-}1\;\;\text{ in the range }\,0 \leq x \leq 2\pi$

    Answers: .$\displaystyle \frac{3\pi}{8},\;\frac{7\pi}{8}$ . These are wrong! **

    We are expect to know that: .$\displaystyle \sin270^o \:=\:\sin\frac{3\pi}{2} \:=\:-1$


    We have: .$\displaystyle \sin2x \:=\:-1$

    . . Then: .$\displaystyle 2x \;=\;\frac{3\pi}{2},\;\frac{7\pi}{2},\;\frac{11\pi }{2},\;\frac{15\pi}{2},\;\hdots$

    . . Hence: .$\displaystyle x \;=\;\frac{3\pi}{4},\;\frac{7\pi}{4},\;\frac{11\pi }{4},\;\frac{15\pi}{4},\;\hdots$


    The roots in the given range are: .$\displaystyle x \;=\;\frac{3\pi}{4},\;\frac{7\pi}{4}$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    Maybe the problem was: .$\displaystyle \sin{\color{red}4}x \:=\:-1$

    Even then, their answer is wrong.
    . . There are more answers in that range.


    If the problem is: .$\displaystyle \sin4x \:=\:-1$

    . . then we have: .$\displaystyle 4x \;=\;\frac{3\pi}{2},\;\frac{7\pi}{2},\;\frac{11\pi }{2},\;\frac{15\pi}{2},\;\hdots$


    Therefore: .$\displaystyle x \;=\;\frac{3\pi}{8},\;\frac{7\pi}{8},\;\frac{11\pi }{8},\;\frac{15\pi}{8}$

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