Question

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

Answer 1

\[\tan 3x=-\sqrt{3}\]

Answer 2

thats in Q2 or Q4

Answer 3

tan(60)

Answer 4

so x = 20, 40 are 2 of em

Answer 5

20 aint right lol

Answer 6

2pi/9

Answer 7

5pi/9

Answer 8

120pi/(180)3 and 300pi/(180)3 maybe lol

Answer 9

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