# trig

posted by on .

(3 tan^2x-1)(tan^2-3)= 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.....

• trig - ,

(3 tan^2x-1)(tan^2-3)= 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 30-60-90 triangle we know that
tan 30°
so in degrees, x = 30°, 150°, 210° and 330°

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"

so π/6 = 30° , by dividing both sides of the equation above by 6
further:
150° = 5π/6
210° = 7π/6
330° = 11π/6

• trig - ,

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

• my bad! - trig - ,

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.