1. ## Solving trigonometric equations

Hello everyone, I have a question

solve for all degrees and theta if 0 degrees <= theta < 360 degrees

1. 2sinTHETA=1

I how its 1/2 and I know they used the unit circle for the answer the answer is 30 + 360k, 150 + 360k(the first part). I want to know were the 360 came from. Why do I need it in my answer? and how can I use this to find out cot, tan, sec, csc, cos?

2. sin function is a periodic function that its period is $2\pi$ it means after $2\pi$ intervals the curve will be repeated and sin of the angles also will be reapted after one cycle of $2\pi$ was completed; and after n iterval it can be calculted like this:

$\sin(n2\pi + \theta) = \sin(\theta)$

3. ## hope this can help

2sin(x)=1
sin(x)=1/2
so your ref angle is 30 or pi/6
becuase the question give you a domain for the value of x. you can only find answer within this domain. In your case is 1 positive full rotation.

sin in Q1 and Q2 have + value.
x=30, and x=150

sin , cos and tan have two + values in 1 positive full rotation.
sin in Q1 and Q2
cos in Q1 and Q4
tan in Q1 and Q3

you only can have x=30. Not x=150, because it is over the domain.

4. Oh now I see it. Thanks. But I still don't understand why I have to but a "+ 360" at the end of it.

5. Originally Posted by Sherina
Oh now I see it. Thanks. But I still don't understand why I have to but a "+ 360" at the end of it.
Ok think of it like this. sin30 and and sin150 equals 1/2

Now when you add 360k, k being all intergers, your just making a full roatation.

For example when k is equal to 1 for

$\sin{(30 + 360(1))}= \frac{1}{2}$

Now try k equals 2 and see what you get.

It has to do with reference angle

6. Originally Posted by 11rdc11
Ok think of it like this. sin30 and and sin150 equals 1/2

Now when you add 360k, k being all intergers, your just making a full roatation.

For example when k is equal to 1 for

$\sin{(30 + 360(1))}= \frac{1}{2}$

Now try k equals 2 and see what you get.

It has to do with reference angle
What is a reference angle?