trig

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(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°

    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

  • 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

    pi/6 radians (30 degrees) is only one of the answers.

  • 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.

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