How to do this?

[allegansveritatem]

Here is the problem:

Here is what I tried to do..it is pretty gibberishy because I seem to start here in medias res, but actually this is just one part of my endeavors and it is presented here merely to give some idea what I tried:

The key to the problem seems to be to translate tan into some form of cos that will then allow me to transform the left side into factors that can be zeroed. No?
PS: This is not an identity but an equation and I am asked to find the solutions for the interval (0,2 pi)

 

[lev888]

Here is the problem:View attachment 18216

Here is what I tried to do..it is pretty gibberishy because I seem to start here in medias res, but actually this is just one part of my endeavors and it is presented here merely to give some idea what I tried:View attachment 18217
The key to the problem seems to be to translate tan into some form of cos that will then allow me to transform the left side into factors that can be zeroed. No?

This is not a problem, this is an equality, or an equation. Can you post the full text?
And if you think it's "pretty gibberishy", why not rewrite it and post something that makes more sense?

 

[topsquark]

The first thing to say is that it can be much easier to write the problem out without putting in those pesky [math]\theta[/math]s, but it is a very sloppy habit. It's a good thing to get used to always putting in the argument.

The idea is to get the equation into a form that only uses one trig function. I usually solve for [math]sin( \theta )[/math] but you can use whatever you would like.
[math]cos( 2 \theta ) - tan( \theta ) = 1[/math]
[math](1 - 2 ~ sin^2( \theta ) ) - \dfrac{sin( \theta )}{\sqrt{ 1 - sin^2( \theta ) }} = 1[/math]
Now let [math]u = sin( \theta )[/math]
[math](1 - 2 u^2) - \dfrac{u}{ \sqrt{1 - u^2}} = 1[/math]
It's messy, but not too awful bad. (Hint: Get rid of the 1's next.)

-Dan

 

[Steven G]

You do realize that sin, cos and tan have no meaning at all. You really have to include angles. In my option to get help on this site you need to write real math--mainly so you learn how to write math.

 

[allegansveritatem]

The first thing to say is that it can be much easier to write the problem out without putting in those pesky [math]\theta[/math]s, but it is a very sloppy habit. It's a good thing to get used to always putting in the argument.

The idea is to get the equation into a form that only uses one trig function. I usually solve for [math]sin( \theta )[/math] but you can use whatever you would like.
[math]cos( 2 \theta ) - tan( \theta ) = 1[/math]
[math](1 - 2 ~ sin^2( \theta ) ) - \dfrac{sin( \theta )}{\sqrt{ 1 - sin^2( \theta ) }} = 1[/math]
Now let [math]u = sin( \theta )[/math]
[math](1 - 2 u^2) - \dfrac{u}{ \sqrt{1 - u^2}} = 1[/math]
It's messy, but not too awful bad. (Hint: Get rid of the 1's next.)

-Dan

I think you are maybe looking to construct some kind of quadratic transaction? Well, I will work along the lines you suggest and we shall see. Thanks

 

[allegansveritatem]

You do realize that sin, cos and tan have no meaning at all. You really have to include angles. In my option to get help on this site you need to write real math--mainly so you learn how to write math.

I will mend my ways in this regard.

 

[allegansveritatem]

This is not a problem, this is an equality, or an equation. Can you post the full text?
And if you think it's "pretty gibberishy", why not rewrite it and post something that makes more sense?

Usually that is what I do and it wasn't until I looked at the photo I had taken did I realize how bad my bit of work looked. By then it was too late, as I did not access to the text or my notebook. As for the expression...When I first posted this I neglected to mention that the instruction for this problem is to find all solutions to the equation in the interval (0,2pi). When it occured to me that I had forgotten to make this important specification, I came back and amended my original post.

 

[allegansveritatem]

The first thing to say is that it can be much easier to write the problem out without putting in those pesky [math]\theta[/math]s, but it is a very sloppy habit. It's a good thing to get used to always putting in the argument.

The idea is to get the equation into a form that only uses one trig function. I usually solve for [math]sin( \theta )[/math] but you can use whatever you would like.
[math]cos( 2 \theta ) - tan( \theta ) = 1[/math]
[math](1 - 2 ~ sin^2( \theta ) ) - \dfrac{sin( \theta )}{\sqrt{ 1 - sin^2( \theta ) }} = 1[/math]
Now let [math]u = sin( \theta )[/math]
[math](1 - 2 u^2) - \dfrac{u}{ \sqrt{1 - u^2}} = 1[/math]
It's messy, but not too awful bad. (Hint: Get rid of the 1's next.)

-Dan

So I went back at it today and came up, after a hard fight, with this:


Yes? No?

 

[Deleted member 4993]

You do realize that sin, cos and tan have no meaning at all. You really have to include angles. In my option to get help on this site you need to write real math--mainly so you learn how to write math.

Do not say that in front of tanning-bed manufacturers!!

 

[Deleted member 4993]

So I went back at it today and came up, after a hard fight, with this:
View attachment 18250

Yes? No?

You can easily check your answers by evaluating your given expression in the equation.

Why are you restricting your domain to 0 to 2pi?

 

[allegansveritatem]

You can easily check your answers by evaluating your given expression in the equation.

Why are you restricting your domain to 0 to 2pi?

That is a restriction imposed by the author of the equations.

 

[allegansveritatem]

You can easily check your answers by evaluating your given expression in the equation.

Right. Also, I can make the original equation into a function and enter it into my graphing calculator, which I have done and I think I am good to go here.