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Math Help - inverse trigonometric functions

  1. #1
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    inverse trigonometric functions

    just need some help getting started

    1.) sin2x= -1/2
    2.) 4cos^2 x-8cosx+1=0
    3.) sin2x= -sinx
    greatly appreciate it
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  2. #2
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    Quote Originally Posted by zukywich08 View Post
    just need some help getting started

    1.) sin2x= -1/2
    2.) 4cos^2 x-8cosx+1=0
    3.) sin2x= -sinx
    greatly appreciate it
    (1) sin(u) = -\frac{1}{2}

    u = \frac{7\pi}{6} \, , \, \frac{11\pi}{6} and those angles coterminal with these two.

    (2) quadratic formula ...

    (3) 2\sin{x}\cos{x} + \sin{x} = 0

    factor and solve
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  3. #3
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    quick question

    for the third problem how would you factor it? do you take sinx out of it?
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  4. #4
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    Quote Originally Posted by zukywich08 View Post
    for the third problem how would you factor it? do you take sinx out of it?
    yes
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  5. #5
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    i got sinx(sinxcosx+1)=0
    but i dont know what to do with the cosine?
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  6. #6
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    Quote Originally Posted by zukywich08 View Post
    i got sinx(sinxcosx+1)=0
    but i dont know what to do with the cosine?
    review your algebra ...

    2\sin{x}\cos{x} + \sin{x} = 0

    \sin{x}(2\cos{x} + 1) = 0

    set each factor = 0 and solve
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  7. #7
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    alright thanks for the help man it came in handy
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