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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\tan 3x=\sqrt{3}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so x = 20, 40 are 2 of em

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0120pi/(180)3 and 300pi/(180)3 maybe lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\text{Tan}[3x]= \sqrt{3}\]\[\text{ArcTan}[\text{Tan}[3x]]=\text{ArcTan}\left[\sqrt{3}\right] \]\[3 x=\frac{\pi }{3}\]\[x=\frac{\pi }{9}\]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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