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Math Help - yr11 2U trig

  1. #1
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    yr11 2U trig

    for 0<= θ <=360, solve (sin θ +1)(2cos θ -1)=0
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by deej813 View Post
    for 0<= θ <=360, solve (sin θ +1)(2cos θ -1)=0
    Applying the zero product property, we have either

    \sin\theta=-1 or \cos\theta=\tfrac{1}{2}.

    Can you take it from here?
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  3. #3
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    do i just put sin^-1(-1) in my calculator
    and cos^-1(1/2)
    ??
    sorry im very confused
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by deej813 View Post
    do i just put sin^-1(-1) in my calculator
    and cos^-1(1/2)
    ??
    sorry im very confused
    You could...but its best if you try to memorize these values from the unit circle: Unit circle - Wikipedia, the free encyclopedia

    In our case, \sin^{-1}\left(-1\right) has only one value in the specified interval for \theta, and its \theta=270^{\circ}

    However, \cos^{-1}\left(\tfrac{1}{2}\right) has two solutions in the given interval for \theta. They are \theta_1=60^{\circ} and \theta_2=300^{\circ}.

    Your final answer to the original question is all three values for \theta.

    Does this make sense?
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  5. #5
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    yes i get it
    finally
    thanks
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