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anonymous

  • one year ago

tanA+sqart3=0,sinA-1/2=0,cosA+squrt3/2,secA-1=0 . which one does not have solution in qurdrant 2 ?

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  1. anonymous
    • one year ago
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    i know i have to use the cast rule but still confused

  2. Michele_Laino
    • one year ago
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    first equation: \[\tan A = - \sqrt 3 \] we have the subsequent drawing: |dw:1434998122110:dw|

  3. Michele_Laino
    • one year ago
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    as we can see one solution belongs to the II quadrant, and the oter one belongs to the fourth quadrant

  4. Michele_Laino
    • one year ago
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    secon equation: \[\sin A = \frac{1}{2}\] |dw:1434998289606:dw| here we have one solution which belongs to the first quadrant, and the second one which belongs to the second quadrant

  5. anonymous
    • one year ago
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    what about the cast rule ?

  6. Michele_Laino
    • one year ago
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    third equation: \[\cos A = - \frac{{\sqrt 3 }}{2}\] |dw:1434998435741:dw| here we have one solution which belongs to the II quadrant, and the second one which belongs to the III quadrant

  7. Michele_Laino
    • one year ago
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    finally, fourth equation: \[\sec A = 1 \to \frac{1}{{\cos A}} = 1 \to \cos A = 1\] |dw:1434998618085:dw| here we have 2 solutions, namely the first one which is A=0, and then it belongs to the I quadrant, whereas the second one, which is A=360, belongs to the IV quadrant. So what can you conclude?

  8. anonymous
    • one year ago
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    do u know how to do this without using the unit circle ?

  9. anonymous
    • one year ago
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    my teacher did the cast rule and it was much faster , kinda forgot though

  10. Michele_Laino
    • one year ago
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    yes! I know, we have to solve all of your four equations. Nevertheless I think that a graphical solution is the best solution

  11. Michele_Laino
    • one year ago
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    sorry, what is the cast rule?

  12. anonymous
    • one year ago
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    btw do u think the solution is thre last one ?

  13. Michele_Laino
    • one year ago
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    yes! That's right!

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