# Thread: Find all values of θ in the interval [0, 2pi]

1. ## Find all values of θ in the interval [0, 2pi]

Find all values of θ in the interval [0, 2pi], given:

cot θ = -1/√3

cosθ= 1/2

I don't really need the answer, just need to know how to solve this type of problem

I greatly appreciate your answer

Thank you.

2. ## Re: Find all values of θ in the interval [0, 2pi]

First, it helps to write the functions in terms of sine, cosine or tangent. Once you have done that, make note of whether your function is positive or negative, take note of which quadrants that function is positive or negative in. Work out the focus angle (for the first quadrant) and then adjust according to whichever quadrants you need to be in.

See how you go.

3. ## Re: Find all values of θ in the interval [0, 2pi]

Originally Posted by Prove It
First, it helps to write the functions in terms of sine, cosine or tangent. Once you have done that, make note of whether your function is positive or negative, take note of which quadrants that function is positive or negative in. Work out the focus angle (for the first quadrant) and then adjust according to whichever quadrants you need to be in.

See how you go.
I can find the first angle but then the second one confuses me, like how do you find the 2nd one?

4. ## Re: Find all values of θ in the interval [0, 2pi]

Well in your first equation, like I said, put it as either a sine, cosine, or tangent equation. The equation is equivalent to $\displaystyle \tan{(\theta)} = -\sqrt{3}$.

In which quadrants is the tangent function negative?

5. ## Re: Find all values of θ in the interval [0, 2pi]

$cot\theta= \frac{1}{tan\theta}$

$tan\theta= \frac{sin\theta}{cos\theta}$

Therefore
$cot\theta= \frac{cos\theta}{sin\theta}$

You know that

$\frac{cos\theta}{sin\theta}=\frac{-1}{\sqrt{3}}$

And
$cos\theta= \frac{1}{2}$

Can you solve these two equations for theta?

6. ## Re: Find all values of θ in the interval [0, 2pi]

Originally Posted by Prove It
Well in your first equation, like I said, put it as either a sine, cosine, or tangent equation. The equation is equivalent to $\displaystyle \tan{(\theta)} = -\sqrt{3}$.

In which quadrants is the tangent function negative?
2 and 4

Originally Posted by Shakarri
$cot\theta= \frac{1}{tan\theta}$

$tan\theta= \frac{sin\theta}{cos\theta}$

Therefore
$cot\theta= \frac{cos\theta}{sin\theta}$

You know that

$\frac{cos\theta}{sin\theta}=\frac{-1}{\sqrt{3}}$

And
$cos\theta= \frac{1}{2}$

Can you solve these two equations for theta?
Yes.
the 1st one theta is 2pi/3
and the 2nd one theta is pi/3

7. ## Re: Find all values of θ in the interval [0, 2pi]

Now you need to try to find ALL solutions in the given region. Like I said, think about which quadrants you need to be in...