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  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    trig solutions

    From (0,\pi)

    how do i solve this?

    2sin(x) + csc(x) = 0

    do i subtract then divide csc(x)?
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  2. #2
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     2\sin(x)+\csc(x)= 0

     2\sin(x)=-\csc(x)

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

    now multiply both sides by sin(x) and divide both sides by 2
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  3. #3
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by pickslides View Post
     2\sin(x)+\csc(x)= 0

     2\sin(x)=-\csc(x)

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

    now multiply both sides by sin(x) and divide both sides by 2
    do i square both sides in the end?
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  4. #4
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    Hello >_<SHY_GUY>_<
    Quote Originally Posted by >_<SHY_GUY>_< View Post
    From (0,\pi)

    how do i solve this?

    2sin(x) + csc(x) = 0

    do i subtract then divide csc(x)?
    Are you sure you have this question right? pickslides has taken you as far as this:
    2\sin x = -\frac{1}{\sin x}
    and has said what to do next. Well, this gives:
    \sin^2x = -\frac12
    In answer to your question 'do I square both sides' - no. You should take the square root now, to get \sin x on the LHS. But can you see what the problem is with trying to take the square root of the RHS?

    As it stands, then, this equation has no real solutions. Perhaps you should check that you copied it down correctly.

    Grandad
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  5. #5
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    Maybe the question was

     2\sin(x) - \csc(x) = 0

    This problem will have real solutions.
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