Analytic Trig

K

Koala

New member
Joined
Mar 17, 2012
Messages
12
Alright, so I'm told to solve each equation on the interval [0, 2pi]

2(sin^2)(theta)-sin(theta)-1=0

Here is what I have done:

Factor out Sin(theta)
Sin(theta)[2(sin(theta))+1]=0

Set each part equal to zero:

Sin(theta)=0 2sin(theta)+1=0
Theta=(Pi), (2Pi) Theta=(7Pi/6), (11Pi/6)

So the solution set would be: {Theta|Theta=(Pi),(2Pi), (4Pi/3), (5Pi/3)}

However, the solutions manual says that the solution set is actually {Theta|Theta=(Pi/2),(7Pi/6),(11Pi/6)}

How do they get that Sin(Theta)=0 means that Theta=(Pi/2)? Because Sin=0 at Pi and 2Pi on the unit circle, which is what I'm working with.

Thanks for taking a look!
 
Koala said:
2 sin^2(theta) - sin(theta) - 1 = 0

Factor out sin(theta)

sin(theta) [2 sin(theta) + 1] = 0
Click to expand...

I stopped reading, at this point, because the factorization is not correct.

We can see that it's not correct because multiplying-out your factored form fails to give back the original expression on the left-hand side.

sin(theta) times [2 sin(theta) + 1] equals:

2 sin^2(theta) + sin(theta)

That doesn't match the exercise.


Instead, you could recognize that the left-hand side is quadratic in form.

You could make a substitution like: let x = sin(theta)

The given equation then becomes:

2x^2 - x - 1 = 0

Do you remember how to factor this quadratic polynomial?

If so, do it, and then switch x in your factors back to sin(theta) and continue from there.

Cheers ~ Mark :cool:
 
Last edited:
Oops! Sorry about the factorization. Correctly factored, it ends up as sin(theta)=0 and sin(theta)=-1/2

So, I end up with the same question... what do I do from here?
 
Koala said:
Any thoughts on what next?
Click to expand...

Sure....you might want to look at a diagram of a unit circle (your text should have one, or they're readily available online) to see WHERE sin @ = 0 (0, pi, 2pi, etc) and where sin @ = -1/2 (I'll let YOU find out where that happens).
 
I know where sin is according to the unit circle. If you read my original post, remember that what I understood was different from the book's answer.

Based on Sin(theta)=0 and Sin(theta)=-1/2

Then Theta should equal Pi, 2Pi, 7Pi/6, and 11Pi/6.

The book says that Theta equals Pi/2, 7Pi/6, and 11Pi/6.

Sin is the y-value... and sin only equals zero when the x-value is Pi or 2Pi (0). How, then, can they say that the answer is at Pi/2 (90 degrees)?
 
Koala said:
I know where sin is according to the unit circle. If you read my original post, remember that what I understood was different from the book's answer.
Click to expand...

You do not need to talk to volunteers like that. If you want to be snippy, please consider paying for the service. Of course, you probably shouldn't do it then, either.

Based on Sin(theta)=0
Click to expand...

Please show your factorization that incorrectly led you to believe that \(\displaystyle \sin(\theta) = 0\) constitutes a possible solution.
 
I'm sorry, I really wasn't trying to be snippy. I hate it when people are snippy, and I do appreciate the volunteers a lot! Thanks for your help, and again, I'm sorry.
 
I'm glad we got past that.

Now, let's move on to the factorization. Something still going wrong, in there.
 
You must log in or register to reply here.
Top