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Hannah_Ahn

  • 4 years ago

Solve algebraically. Express the roots in exact form. 2sinxcosx-1 = 0 the answer is pi over 4 + n pi and I don't know the steps please help, thanks :)

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  1. myininaya
    • 4 years ago
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    sin(2x)-1=0 sin(2x)=1

  2. myininaya
    • 4 years ago
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    when is sin(u), 1?

  3. myininaya
    • 4 years ago
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    when u=pi/2+2npi u=3pi/2+2npi so that means 2x=pi/2+2npi 2x=3pi/2+2npi

  4. Hannah_Ahn
    • 4 years ago
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    pi over 2

  5. myininaya
    • 4 years ago
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    to solve for x just divide both sides by 2

  6. Hannah_Ahn
    • 4 years ago
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    lovely! thanks. have a wonderful night :)

  7. myininaya
    • 4 years ago
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    np

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