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anonymous

  • 5 years ago

solve over the interval [0,2 pi): tan 3x=the negative square root of 3

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  1. anonymous
    • 5 years ago
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    \[\tan 3x=-\sqrt{3}\]

  2. amistre64
    • 5 years ago
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    thats in Q2 or Q4

  3. amistre64
    • 5 years ago
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    tan(60)

  4. amistre64
    • 5 years ago
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    so x = 20, 40 are 2 of em

  5. amistre64
    • 5 years ago
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    20 aint right lol

  6. amistre64
    • 5 years ago
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    2pi/9

  7. amistre64
    • 5 years ago
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    5pi/9

  8. amistre64
    • 5 years ago
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    120pi/(180)3 and 300pi/(180)3 maybe lol

  9. anonymous
    • 5 years ago
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    \[\text{Tan}[3x]= -\sqrt{3}\]\[\text{ArcTan}[\text{Tan}[3x]]=\text{ArcTan}\left[-\sqrt{3}\right] \]\[3 x=-\frac{\pi }{3}\]\[x=-\frac{\pi }{9}\]Problem calls for positive angles between 0 and 2 pi\[x=\frac{8 \pi }{9} \text{and } x=\frac{17 \pi }{9} \]

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