trig
posted by help? .
(3 tan^2x1)(tan^23)= 0
can someone please explain how to do this problem and then explain the whole pie thing at the end...like, this is suppose to equal pie/6 b/c it's tan...im soooo confused.....

(3 tan^2x1)(tan^23)= 0
so 3tan^2x = 1 or tan^2x = 3
tan^2x = 1/3 or tan^2x = 3
the second part has no solution, since something^2 cannot be negative
tanx = ± 1/√3
so x is in any of the 4 quadrants.
form the 306090 triangle we know that
tan 30°
so in degrees, x = 30°, 150°, 210° and 330°
It appears that we are to give the answers in radians, and you seem to know very little about radians.
I am sure your texbook has a unit on it, or else you can learn about it on the thousands of webpages found after googling "radians"
in short, πradians = 180°
so π/6 = 30° , by dividing both sides of the equation above by 6
further:
150° = 5π/6
210° = 7π/6
330° = 11π/6 
Your equation is already factored. You have a solution if either factor is zero.
tan^2x can be either 1/3 or 3
tan x = +or 1/sqrt3
tanx = +or sqrt3
x = 30, 60, 120, 150, 210, 240, 300 or 330 degrees
pi/6 radians (30 degrees) is only one of the answers. 
Go with drwls solution.
somehow my +3 changed into a 3 for the second factor. there are 4 more solutions from that, as drwls shows.