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Math Help - Trigonometric Equations

  1. #1
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    Red face Trigonometric Equations

    Hello

    I have a couple of trig equations which i am having difficulty solving. I was sick for a while before the break and now that its exam review time I don't remember much. BTW this is an assignment due tomorrow, thats why its in urgent help.

    Here are the questions



    3cot^2(3x-\frac{pi}{4})-1=0
    sin2x+sinx-2cosx-1=0
    sinx+\sqrt{3}cosx=1 please note the cos x is not under the root

    heres how i started solving the first equation =\

    3cot^2(3x-\frac{pi}{4})=1
    cot(3x-\frac{pi}{4}=\sqrt{\frac{1}{3}}
    tany=\frac{1}{\sqrt{\frac{1}{3}}}
    =\sqrt{3}
    therefore
    y = \frac{2pi}{3} y = \frac{pi}{3} y = \frac{4pi}{3} y = \frac{5pi}{3}
    now im kinda stuck. since the question does not give a limit (ie 0 to 2pi) do we need to figure out an equation or something? =\



    any help appreciated
    thanks
    chris
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  2. #2
    Super Member wingless's Avatar
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    3\text{Cot}^2\left(3x-\frac{\pi }{4}\right)=1
    \text{Cot}^2\left(3x-\frac{\pi }{4}\right)= \frac{1}{3}

    (Remember that \sqrt{x^2} = |x|)

    \left|\text{Cot}\left(3x-\frac{\pi }{4}\right)\right| = \sqrt{\frac{1}{3}}
    \text{Cot}\left(3x-\frac{\pi }{4}\right)=\pm \sqrt{\frac{1}{3}}
    \text{Cot}\left(3x-\frac{\pi }{4}\right)= +\sqrt{\frac{1}{3}} \text{ }  \boxed{\text{I}} \text{ } , \text{ } \text{Cot}\left(3x-\frac{\pi }{4}\right)=-\sqrt{\frac{1}{3}} \text{ } \boxed{\text{II}}

    \boxed{I.} \text{  } \text{Cot}\left(3x-\frac{\pi }{4}\right)= \sqrt{\frac{1}{3}}
    3x-\frac{\pi }{4}+ \text{k$\pi $} = \frac{\pi }{3}
    x = \frac{7\pi }{36}+\frac{\text{k$\pi $}}{3}

    \boxed{II.} \text{  } \text{Cot}\left(3x-\frac{\pi }{4}\right)=-\sqrt{\frac{1}{3}}
    3x-\frac{\pi }{4} + \text{k$\pi $} = \frac{2\pi }{3}
    x = \frac{11\pi }{36}+\frac{\text{k$\pi $}}{3}

    So,

    x=\{ \frac{7\pi }{36}+\frac{\text{k$\pi $}}{3}, \frac{11\pi }{36}+\frac{\text{k$\pi $}}{3}  \}
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  3. #3
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    ok i think im understanding how you got the answers,
    i changed y back to the original form but how dod you come up with the later part of x = \frac{7\pi }{36}+\frac{\text{k$\pi $}}{3}
    like as in why it is pi / 3
    and also, do the other answers not work? or did you omit them because you get them the same way or they are redundant?
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  4. #4
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    Hello, Chris!

    Here's some help . . .


    \sin2x + \sin x - 2\cos x - 1\;=\;0

    We have: . 2\sin x\cos x + 2\sin x - 2\cos x - 1 \;=\;0

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

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


    Then we have: . 2\cos x + 1\:=\:0\quad\Rightarrow\quad \cos x \:=\:-\frac{1}{2}

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


    Therefore: . x \;=\;\left\{\dfrac{2\pi}{3},\;\dfrac{4\pi}{3},\;\d  frac{\pi}{2} \right\} + 2\pi n




    \sin x+\sqrt{3}\cos x \;=\;1

    This one requires some Olympic-level gymnastics . . . and a few facts:

    . . \sin\frac{\pi}{3} \:=\:\frac{\sqrt{3}}{2},\;\cos\frac{\pi}{3} \:=\:\frac{1}{2} . [1]

    . . \sin A\cos B + \sin B\cos A \:=\:\sin(A + B) . [2]


    We have: . \sin x + \sqrt{3}\cos x \;=\;1

    Divide by 2: . \frac{1}{2}\sin x + \frac{\sqrt{3}}{2}\cos x \;=\;\frac{1}{2}

    From [1], we have: . \cos\frac{\pi}{3}\sin x + \sin\frac{\pi}{3}\cos x \;=\;\frac{1}{2}

    From [2], we have: . \sin\left(x + \frac{\pi}{3}\right) \;=\;\frac{1}{2}

    Hence: . x + \frac{\pi}{3} \:=\:\left\{\frac{\pi}{6},\;\frac{5\pi}{6}\right\} + 2\pi n

    Therefore: . x \;=\;\left\{-\frac{\pi}{6},\;\frac{\pi}{2}\right\} + 2\pi n



    The first problem has about a dozen basic forms.
    I'll try to get it straightened out later.
    .
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