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emilyr123

  • 2 years ago

4sin(theta)+3sqrt3=sqrt3 solve w/ interval 0 < theta < 2pi

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  1. psi9epsilon
    • 2 years ago
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    Pretty straightforward \[4 \sin (\theta) = \sqrt{3} - 3\sqrt{3}\]

  2. psi9epsilon
    • 2 years ago
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    \[4\sin (\theta)=-2\sqrt{3}\]

  3. psi9epsilon
    • 2 years ago
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    \[\sin (\theta)=-2\sqrt{3 } / 4\]

  4. psi9epsilon
    • 2 years ago
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    \[\sin (\theta)=-\sqrt{3} / 2\]

  5. psi9epsilon
    • 2 years ago
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    \[\theta = \sin^{-1} (-\sqrt{3}/2)\]

  6. psi9epsilon
    • 2 years ago
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    sin THETA is NEGATIVE only in 3rd and 4th quadrant

  7. psi9epsilon
    • 2 years ago
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    \[\theta=\sin^{-1} (\sin(\pi+\pi/3))\]

  8. psi9epsilon
    • 2 years ago
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    THETA = 4 PI /3

  9. psi9epsilon
    • 2 years ago
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    OR THETA = 2PI - PI/3 = 5PI/3

  10. psi9epsilon
    • 2 years ago
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    TWO ANSWERS as the interval is from 0 -> 2 Pi

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