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4meisu

  • 2 years ago

Given that 2sin^2 θ + sinθ - 1 = 0, find the two values for sin θ

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  1. ashna
    • 2 years ago
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    2sin² θ + sin θ - 1 = 0 (2sin θ - 1)(sin θ + 1) = 0

  2. lgbasallote
    • 2 years ago
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    imagine \(\sin \theta = x\) so this thingy will become 2x^2 + x - 1 = 0 do you know how to solve for x there?

  3. ashna
    • 2 years ago
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    so much can you understand ?

  4. ashna
    • 2 years ago
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    @4meisu ?

  5. 4meisu
    • 2 years ago
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    No I do not know how to solve for x.. What's next?

  6. ashna
    • 2 years ago
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    Now we can split that into two equations 2sin θ - 1 = 0 and sin θ + 1 = 0 this much understood ?

  7. 4meisu
    • 2 years ago
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    yes

  8. 4meisu
    • 2 years ago
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    sinθ = 1/2 and sinθ = -1 ?

  9. ashna
    • 2 years ago
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    Solving the first one: 2sin θ - 1 = 0 sin θ = 1/2 θ = 30°, 150°

  10. ashna
    • 2 years ago
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    Second: sin θ + 1 = 0 sin θ = -1 θ = 270°

  11. 4meisu
    • 2 years ago
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    Okay thank you

  12. ashna
    • 2 years ago
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    :)

  13. 4meisu
    • 2 years ago
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    The question is asking for only two values for theta though, which one would we choose?

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