Question

Solve the equation on the interval 0≤θ<2π:
tan(θ/2)=((√3)/3) Explain please!

Answer 1

did you try ?

Answer 2

I started with trying to separate tan(theta) from the rest of it, is that right?

Answer 3

can you simplify this : ((√3)/3) ?

Answer 4

I don't think so

Answer 5

\(\sqrt a/a=\sqrt a/[(\sqrt a)(\sqrt a)]= 1/ \sqrt a\)
did you get this ?

Answer 6

So it would be 1/√3?

Answer 7

yes, for what values of theta is tan value 1/sqrt 3 ??

Answer 8

But wait, where did a or the denominator go

Answer 9

'a' was just an example to show you that \(\sqrt3/3=1/\sqrt 3\)
and which denominator ?

Answer 10

The denominator of √3/3 o:

Answer 11

denominator of √3/3 =3
but the whole √3/3 equals, 1/√3
so, tan (theta /2) = 1/ √3
for what values of angle is tan value 1/sqrt 3

Answer 12

Oy okay. And I'm not sure o; do I have to solve out the different points on the unit circle to get it?

Answer 13

tan = sin / cos
so, see for which angle, this ratio comes out to be 1/sqrt 3

Answer 14

π/3 and 4π/3?

Answer 15

ok, range of theta is 0 to 2pi
so, range of theta/2 (because thats our angle) will be 0 to pi/2
hence, we select only pi/3
theta /2 = pi/3
theta = .... ?

Answer 16

range of theta/2 (because thats our angle) will be 0 to pi *****

Answer 17

π/3? o:

Answer 18

theta /2 = pi/3
theta = 2pi/3 o:

Answer 19

And that's the answer? o;

Answer 20

yes, thats it.

Answer 21

any doubts ?

Answer 22

My teacher put down that the answer was π/3, did I mess up somewhere? o:

Answer 23

oh, yes :P the angles are not π/3 and 4π/3 for tan value of 1/sqrt 3
i didn't check/verify, assumed you are correct :P

Answer 24

its pi/6 in the range 0 to pi
so, theta /2 = pi/6
theta = 2pi/6 = pi/3

Answer 25

sin pi/6 = 1/2
cos pi/6 = sqrt 3/ 2
tan pi/6 = sin pi/6 / cos pi/6 = 1/ sqrt 3

Answer 26

Oh okay, I got it! Thank you very much (:

Answer 27

welcome ^_^