Question

help plz boys.
Consider the equation 3tan^2(x)-1=0
The solutions of this equation over the interval [0, π] are
a) 2pi/3, 5pi/6
b) pi/6, 2pi/3
c) 7pi/6, 11pi/6
d) pi/6, 5pi/6

plz help

Answer 1

What do you get when solving for tan(x) ?

Answer 2

I get \[\frac{\sqrt{3}}{3}\]

Answer 3

Should be 1/sqrt(3)

Answer 4

or -1/sqrt(3)

Answer 5

yes, but rationalize the denominator :)

Answer 6

Lol I didn't notice :P

Answer 7

But you forgot about -sqrt(3)/3

Answer 8

right .. you got me ..., plus or minus

Answer 9

Yeah, after that you just apply the inverse function in a calculator..

Answer 10

It's called arctan

Answer 11

ot (-5pi/12)
well that will be in quadrant 4 since it is clockwise from the x axis due to the minus sign and 5 pi/12 is less than pi/2 or 6 pi/12.
call the x component of this vector a and the y component b. Then the magnitude of the cotan is a/b.
Notice that the angle to the negative y axis is 6pi/12 - 5pi/12 = pi/12
then the magnitude of tan pi/12 is a/b, just what we are looking for
so what is tan pi/12?
In degrees it is 15 degrees.
We know tan 30 degrees = 1/sqrt3 and sin 30 =1/2 and cos 30 = sqrt3 /2
so what is tan 15 degrees?
well
tan 15 = sin 30/(1+cos 30)
I think you can take it from there.

Answer 12

what... calculator.... just use the unit circle.
\[ \frac{\sqrt{3}}{3} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\]
so when is sin(x) =1/2

Answer 13

noticed @wibrahim is offline...so I guess this is just us

Answer 14

Yeah, we'll continue though for future reference ;]

Answer 15

I'm getting pi/6 and 5pi/6

Answer 16

Yep, correct !