## jkl3inberg 2 years ago Find all solutions of the equation in the interval [0,2pi). Write your answer in terms of pi. √3 cot(theta)-1=0 I got cot(theta)= pi/3, but the identity of cotangent is x/y, so I'm not sure where to look at. Is the answer just pi/3?

1. hartnn

did you mean you got thetha = pi/3 ?

2. NoelGreco

That's one answer, but there is another angle on that interval where the cotangent is positive and the reference angle is also pi/3.

3. jkl3inberg

Yes hartnn, and NoelGreco the only place there's a reference angle of pi/3 is 1/2 and root3/2, are those my answers?

4. hartnn

he meant there are 2 solutions you got cot theta =1/ sqrt 3 right ? there are 2 angles in [0,2pi) which satisfies this, one angle is pi/3 which you got it correct and other will be (pi/3 +pi) =... ? because cot function is periodic in pi

5. jkl3inberg

I got 1/sqrt 3 yes, but that needs to be rationalized to root 3 / 3, right? Which I don't see anywhere on the unit circle

6. hartnn

oh, you are using unit circle, ok then cot = cos / sin search for the values where the ratio of cos/ sin would equal 1/sqrt 3 one when cos = 1/2 and sin = sqrt 3 /2 and one more when cos =-1/2 and sin = -sqrt 3/2 those 2 angles will be your solution

7. jkl3inberg

so pi/3 and 4pi/3, thanks. I get these kinds of problems but cot=x/y really confused me. How exactly did you end up with 1/2 and root3/2?

8. hartnn

ok, cot (theta) = 1/ sqrt 3 is the ratio of cos and sine functions so, for sin and cos the standard values are 1/2 and sqrt 3/2 for the standard angle pi/3, (we need to remember standard angles and their values or we can use the unit circle) ratio of only these would give us 1/sqrt 3...thats how.

9. jkl3inberg

Ok, got it. Thanks!

10. hartnn

welcome ^_^