having trouble

[allegansveritatem]

Here is the equation that I am asked to find all solutions for within the interval of (0,2pi)


Here is one of many attempts I have made to solve this: part 1:

part 2:


It is crazy wrong but....can anyone give me a pointer here toward the light?
I find that as I go deeper into trigonometry I get more mired in difficulty. I am going very slowly, not rushing, but still it is tough going. I think part of it is I get tense and really this stuff demands being willing to play with a lot of options.

 

[JeffM]

I'd start by reducing this to a function made up of just one trigonometric function in the same variable, namely t.

[MATH]tan(2t) - 2 cos(t) = 0 \implies \dfrac{sin(2t)}{cos(2t)} = 2 cos(t) \implies[/MATH]
[MATH]\dfrac{2sin(t)cos(t)}{1 - 2 sin^2(t)} = 2 cos(t) \implies \dfrac{sin(t)}{1 - 2 sin^2(t)} = 1 \implies[/MATH]
[MATH]sin(t) = 1 - 2 sin^2(t) \implies 2sin^2(t) + sin(t) - 1 = 0.[/MATH]
Now what?

 

[topsquark]

I'd start by reducing this to a function made up of just one trigonometric function in the same variable, namely t.

[MATH]tan(2t) - 2 cos(t) = 0 \implies \dfrac{sin(2t)}{cos(2t)} = 2 cos(t) \implies[/MATH]
[MATH]\dfrac{2sin(t)cos(t)}{1 - 2 sin^2(t)} = 2 cos(t) \implies \dfrac{sin(t)}{1 - 2 sin^2(t)} = 1 \implies[/MATH]
[MATH]sin(t) = 1 - 2 sin^2(t) \implies 2sin^2(t) + sin(t) - 1 = 0.[/MATH]
Now what?

A slight addition: JeffM divided both sides by cos(t). That means one possible solution might be cos(t) = 0.

-Dan

 

[JeffM]

A slight addition: JeffM divided both sides by cos(t). That means one possible solution might be cos(t) = 0.

-Dan

Off to join Subhotosh in the corner!

 

[allegansveritatem]

I'd start by reducing this to a function made up of just one trigonometric function in the same variable, namely t.

[MATH]tan(2t) - 2 cos(t) = 0 \implies \dfrac{sin(2t)}{cos(2t)} = 2 cos(t) \implies[/MATH]
[MATH]\dfrac{2sin(t)cos(t)}{1 - 2 sin^2(t)} = 2 cos(t) \implies \dfrac{sin(t)}{1 - 2 sin^2(t)} = 1 \implies[/MATH]
[MATH]sin(t) = 1 - 2 sin^2(t) \implies 2sin^2(t) + sin(t) - 1 = 0.[/MATH]
Now what?

that is an interesting little stunt you pull when in the second line you cancel out the cos and the 2 from the numerator on LS and replacing it with a 1 on the RS. I like that! I wasn't aware of that one, but yes, it can be done. I guess I should have learned that in arithmetic class before the flood, but better late than never. Yes, I will cehck out your lead. Thanks.
PS: Now that I think of it, the reason I caught on to what happened in the second line is that with all this fooling I h ave been doing with trig equations etc, I have had to build a lot of little models to check if such and such a procedure would fly and I seem to recall having come upon the fact that it is possilbe to divide through a fraction, so to speak, as you did here.

 

[allegansveritatem]

A slight addition: JeffM divided both sides by cos(t). That means one possible solution might be cos(t) = 0.

-Dan

by 2 cos t, no? I will have to think about what you are saying this might mean. I confess, I don't see why it sould mean 0 is a possible solution.

 

[JeffM]

by 2 cos t, no? I will have to think about what you are saying this might mean. I confess, I don't see why it sould mean 0 is a possible solution.

Here is the thing.

In my little stunt as you called it, I was ASSUMING IMPLICITLY that cos(t) was not zero because we cannot divide by zero in the real numbers. So if zero is a solution, performing that stunt will miss it. Therefore, before doing the stunt, you must test to see whether or not zero is a solution. Only after doing that can you use my stunt to look for non-zero solutions. Here is an example where it is obvious.

[MATH]x^6 - 16x^4 = 0 \implies x^2 - 16 = 0 \implies x = \pm 4.[/MATH]
THAT IS AN ERROR.

We missed the solution x = 0. The correct way is

[MATH]x = 0 \text { or } x \ne 0. \\ 0^6 - 16 * 0^4 = 0 - 0 = 0 \implies x = 0 \text { is a solution.}\\ x \ne 0 \text { and } x^6 - 16x^4 = 0 \implies \\ \dfrac{x^6 - 16x^4}{x^4} = \dfrac{0}{x^4} \implies x^2 - 16 = 0 \implies x = \pm 4. \\ \therefore x = - 4, \ x = 0, \text { or } x = 4.[/MATH]
So Dan was quite correctly pointing out in a very kind way that I was guilty of a major error.

 

[JeffM]

To clarify my previous post, dividing on the assumption that the solution is not zero does not mean that zero is a solution. It means that you may miss a solution that is zero. So you must always test whether or not zero is a solution before doing the division.

This is different from arithmetic where you are dividing by a known quantity.

 

[allegansveritatem]

To clarify my previous post, dividing on the assumption that the solution is not zero does not mean that zero is a solution. It means that you may miss a solution that is zero. So you must always test whether or not zero is a solution before doing the division.

This is different from arithmetic where you are dividing by a known quantity.

thanks for clairifying. Actually, I don't think zero was a solution---I entered the equation in my graphing calculator and zero was not broached, as shown here:


Here is what I got when working out problem. Can you say why the calculator image has a solution at pi/2?


Please explain a little further why dividing by 2 cos x needs to be restricted in the way you indicate. This is not at all clear to me.

 

[JeffM]

We have the following very general situation, and we are trying to solve for x.

[MATH]f(x) * g(x) = f(x) * h(x)[/MATH]
There are exactly two possibilities.

[MATH]\text {there exist one or more values at which } f(x) = 0 \text { or}\\ \text {there do not exist any values at which } f(x) = 0.[/MATH]In the first case, each of those values is a solution to our general problem because it is certainly true that

[MATH]0 * g(x) = 0.[/MATH]
So the first thing to do is to try to solve f(x) = 0. It is NOT because f(x) must have one or more x-intercepts. Not at all. It is because f(x) MAY have one or more x-intercepts, and if it does, each is a valid solution to our problem.

With me to here?

Now we can say

[MATH]\text {Excluding any value for which } f(x) = 0,\\ f(x) * g(x) = f(x) * h(x) \implies\\ \dfrac{f(x) * g(x)}{f(x)} = \dfrac{f(x) * h(x)}{f(x)} \implies g(x) = h(x).[/MATH]We are permitted to do the division because we have excluded the possibility that f(x) = 0, and, because we have already checked for solutions in that case, we will not overlook them. Moreover, the division simplifies the problem, and we now look for solutions to g(x) = h(x) because it is certainly true that

[MATH]\text {IF } g(x) = h(x), \text { then } f(x) * g(x) = f(x) * h(x).[/MATH]
For this problem, h(x) = 1, but the technique holds in far more complicated cases. But to ensure that you are not missing any valid solutions, you must check for any cases where f(x) (in the case of this problem 2 cos t) equals zero.

Have it now?

Let's go back to the original problem

[MATH]tan(2t) - 2 cos (t) = 0 \implies tan(2t) = 2 cos(t).[/MATH]
In the specified interval, [MATH]2cos(t) = 0 \text { at } t = \dfrac{\pi}{2} \text { and } \dfrac{3 \pi}{2}.[/MATH]
Those are solutions because

[MATH]tan \left (2 * \dfrac{\pi}{2} \right ) = tan (\pi) = 0 \text { and } 0 - 0 = 0; \text { and }\\ tan \left (2 * \dfrac{3\pi}{2} \right ) = tan (3\pi) = 0 \text { and } 0 - 0 = 0.[/MATH]

 

[allegansveritatem]

We have the following very general situation, and we are trying to solve for x.

[MATH]f(x) * g(x) = f(x) * h(x)[/MATH]
There are exactly two possibilities.

[MATH]\text {there exist one or more values at which } f(x) = 0 \text { or}\\ \text {there do not exist any values at which } f(x) = 0.[/MATH]In the first case, each of those values is a solution to our general problem because it is certainly true that

[MATH]0 * g(x) = 0.[/MATH]
So the first thing to do is to try to solve f(x) = 0. It is NOT because f(x) must have one or more x-intercepts. Not at all. It is because f(x) MAY have one or more x-intercepts, and if it does, each is a valid solution to our problem.

With me to here?

Now we can say

[MATH]\text {Excluding any value for which } f(x) = 0,\\ f(x) * g(x) = f(x) * h(x) \implies\\ \dfrac{f(x) * g(x)}{f(x)} = \dfrac{f(x) * h(x)}{f(x)} \implies g(x) = h(x).[/MATH]We are permitted to do the division because we have excluded the possibility that f(x) = 0, and, because we have already checked for solutions in that case, we will not overlook them. Moreover, the division simplifies the problem, and we now look for solutions to g(x) = h(x) because it is certainly true that

[MATH]\text {IF } g(x) = h(x), \text { then } f(x) * g(x) = f(x) * h(x).[/MATH]
For this problem, h(x) = 1, but the technique holds in far more complicated cases. But to ensure that you are not missing any valid solutions, you must check for any cases where f(x) (in the case of this problem 2 cos t) equals zero.

Have it now?

Let's go back to the original problem

[MATH]tan(2t) - 2 cos (t) = 0 \implies tan(2t) = 2 cos(t).[/MATH]
In the specified interval, [MATH]2cos(t) = 0 \text { at } t = \dfrac{\pi}{2} \text { and } \dfrac{3 \pi}{2}.[/MATH]
Those are solutions because

[MATH]tan \left (2 * \dfrac{\pi}{2} \right ) = tan (\pi) = 0 \text { and } 0 - 0 = 0; \text { and }\\ tan \left (2 * \dfrac{3\pi}{2} \right ) = tan (3\pi) = 0 \text { and } 0 - 0 = 0.[/MATH]

so,what you are saying is this: before you try to divide something into something, make sure the divisor is not or ,cannot be = to zero? And 2cos t can = zero here? That is where the mrub is for me. I mean, tan(2t) - 2cos(t) doesn't have to equal zero just because 2cos(t) = zero. I am pretty dense and can't see how to prove 2cos (t) can equal zero...but I guess it can provided tan (2t) also equals zero. But this would hold then, for any equation of this type, ,any equation with two terms separated by a minus sign, that equals zero, no? I am slow-brained, especially at this small hour of the morning. I need to think this thing through, which I certainly will on the morrow. Thanks for setting it all out. I will post the results of my cogitations.

 

[topsquark]

This is probably the better way to look at it:

In the real number system if ab = 0 then either a = 0 or b = 0.

If we have
[math]\dfrac{2 ~ sin(t) ~ cos(t)}{1 - 2 ~ sin^2(t)} = 2 ~ cos(t)[/math]
[math]\dfrac{2 ~ sin(t) ~ cos(t)}{1 - 2 ~ sin^2(t)} - 2 ~ cos(t) = 0[/math]
[math]\left ( \dfrac{sin(t)}{1 - 2 ~ sin^2(t)} - 1 \right ) (2 ~ cos(t) ) = 0[/math]
Thus either
[math]\dfrac{sin(t)}{1 - 2 ~ sin^2(t)} - 1 = 0[/math]or
[math]2 ~ cos(t) = 0[/math]
-Dan

 

[allegansveritatem]

We have the following very general situation, and we are trying to solve for x.

[MATH]f(x) * g(x) = f(x) * h(x)[/MATH]
There are exactly two possibilities.

[MATH]\text {there exist one or more values at which } f(x) = 0 \text { or}\\ \text {there do not exist any values at which } f(x) = 0.[/MATH]In the first case, each of those values is a solution to our general problem because it is certainly true that

[MATH]0 * g(x) = 0.[/MATH]
So the first thing to do is to try to solve f(x) = 0. It is NOT because f(x) must have one or more x-intercepts. Not at all. It is because f(x) MAY have one or more x-intercepts, and if it does, each is a valid solution to our problem.

With me to here?

Now we can say

[MATH]\text {Excluding any value for which } f(x) = 0,\\ f(x) * g(x) = f(x) * h(x) \implies\\ \dfrac{f(x) * g(x)}{f(x)} = \dfrac{f(x) * h(x)}{f(x)} \implies g(x) = h(x).[/MATH]We are permitted to do the division because we have excluded the possibility that f(x) = 0, and, because we have already checked for solutions in that case, we will not overlook them. Moreover, the division simplifies the problem, and we now look for solutions to g(x) = h(x) because it is certainly true that

[MATH]\text {IF } g(x) = h(x), \text { then } f(x) * g(x) = f(x) * h(x).[/MATH]
For this problem, h(x) = 1, but the technique holds in far more complicated cases. But to ensure that you are not missing any valid solutions, you must check for any cases where f(x) (in the case of this problem 2 cos t) equals zero.

Have it now?

Let's go back to the original problem

[MATH]tan(2t) - 2 cos (t) = 0 \implies tan(2t) = 2 cos(t).[/MATH]
In the specified interval, [MATH]2cos(t) = 0 \text { at } t = \dfrac{\pi}{2} \text { and } \dfrac{3 \pi}{2}.[/MATH]
Those are solutions because

[MATH]tan \left (2 * \dfrac{\pi}{2} \right ) = tan (\pi) = 0 \text { and } 0 - 0 = 0; \text { and }\\ tan \left (2 * \dfrac{3\pi}{2} \right ) = tan (3\pi) = 0 \text { and } 0 - 0 = 0.[/MATH]

I went to bed and slept a short time and woke to the call of nature. I answered said call and on my way back to bed BOOM I saw a great light: If you have two terms separated by a minus sign followed by an = sign and a zero, that means that the two terms have the same value! From there on the rest was easy. Thanks again for the explanation.

 

[JeffM]

First, please read Dan's excellent response. But I think one reason you have may be having trouble here is because of a blind spot on the specifics of this problem..

so,what you are saying is this: before you try to divide something into something, make sure the divisor is not or ,cannot be = to zero?

This is always true, not just with respect to the specific topic we are discussing. And if you want to divide by a variable you either have to make sure in advance that it cannot be zero or you have to proceed by cases, case 1 not dividing when the variable is zero and case 2 dividing when it is not.

And 2cos t can = zero here? That is where the mrub is for me. I mean, tan(2t) - 2cos(t) doesn't have to equal zero just because 2cos(t) = zero.

Oh but it certainly does. First, let's note that the cosine function can equal zero in the interval you are concerned with.

[MATH]cos(t) = 0 \iff t = \dfrac{(2k - 1) \pi}{2} \text { and } k \in \mathbb Z.[/MATH]
You know that perfectly well. The cosine function has an infinite number of x-intercepts, and they are periodically placed. In your specific problem, the argument is confined to the interval (0, 2 * pi). In that interval, there are exactly two values of t where the cosine function is zero, namely

[MATH]cos \left ( \dfrac{\pi}{2} \right ) = 0 \text { and } cos \left ( \dfrac{3 \pi}{2} \right ) = 0 \implies \\ 2cos \left ( \dfrac{\pi}{2} \right ) = 0 \text { and } 2cos \left ( \dfrac{3 \pi}{2} \right ) = 0[/MATH]The tangent function also has an infinite number of x-intercepts, and they too are periodically placed.

[MATH]tan(t) = 0 \iff t = m \pi \text { and } m \in \mathbb Z.[/MATH]
Now for your proof

[MATH]cos(t) = 0 \implies \exists \text { an integer } k \text { such that } t = \dfrac{(2k - 1) \pi}{2} \implies\\ 2t = \cancel 2 * \dfrac{(2k - 1) \pi}{\cancel 2} = (2k - 1) \pi\\ \text {But } k \text { is an integer } \implies (2k - 1) \text {is an integer} \implies\\ tan(2t) = 0.[/MATH]In fact, [MATH]cos(t) = 0 \iff tan(2t) = 0.[/MATH]
I think you are forgetting that the tangent function = sine over cosine and thus is not independent of the sine and cosine. Your concern would be correct if you were dealing with two independent functions, but you are not.

Looking at the specific values of t that make 2cos(t) = 0 for your problem, let's see what the specific values are for 2t and so for tan(2t).

[MATH]t = \dfrac{\pi}{2} \implies 2t = \pi \implies tan(2t) = 0 \text { and}\\ t = \dfrac{3 \pi}{2} \implies 2t = 3 \pi \implies tan(2t) = 0.[/MATH]
You say 3 pi is out of range. No. It is not stipulated that 2t < 2 * pi, just that t < 2 * pi.

am pretty dense and can't see how to prove 2cos (t) can equal zero...but I guess it can provided tan (2t) also equals zero. But this would hold then, for any equation of this type, ,any equation with two terms separated by a minus sign, that equals zero, no? I am slow-brained, especially at this small hour of the morning. I need to think this thing through, which I certainly will on the morrow. Thanks for setting it all out. I will post the results of my cogitations.

You are not dense. Things like independence are fundamental and perhaps not stressed enough to beginning students.

And yes

[MATH]a - b = 0 \iff a = b\\ \therefore a - b = 0 \text { and } b = 0 \implies a = 0.[/MATH]

 

[allegansveritatem]

This is probably the better way to look at it:

In the real number system if ab = 0 then either a = 0 or b = 0.

If we have
[math]\dfrac{2 ~ sin(t) ~ cos(t)}{1 - 2 ~ sin^2(t)} = 2 ~ cos(t)[/math]
[math]\dfrac{2 ~ sin(t) ~ cos(t)}{1 - 2 ~ sin^2(t)} - 2 ~ cos(t) = 0[/math]
[math]\left ( \dfrac{sin(t)}{1 - 2 ~ sin^2(t)} - 1 \right ) (2 ~ cos(t) ) = 0[/math]
Thus either
[math]\dfrac{sin(t)}{1 - 2 ~ sin^2(t)} - 1 = 0[/math]or
[math]2 ~ cos(t) = 0[/math]
-Dan

I am just getting ready to leave but I will get back to this tonight. It looks good. Thanks

 

[allegansveritatem]

This is probably the better way to look at it:

In the real number system if ab = 0 then either a = 0 or b = 0.

If we have
[math]\dfrac{2 ~ sin(t) ~ cos(t)}{1 - 2 ~ sin^2(t)} = 2 ~ cos(t)[/math]
[math]\dfrac{2 ~ sin(t) ~ cos(t)}{1 - 2 ~ sin^2(t)} - 2 ~ cos(t) = 0[/math]
[math]\left ( \dfrac{sin(t)}{1 - 2 ~ sin^2(t)} - 1 \right ) (2 ~ cos(t) ) = 0[/math]
Thus either
[math]\dfrac{sin(t)}{1 - 2 ~ sin^2(t)} - 1 = 0[/math]or
[math]2 ~ cos(t) = 0[/math]
-Dan

Good. I like it! I will give that a shot. Thanks

 

[Cubist]

Thus either
[math]\dfrac{sin(t)}{1 - 2 ~ sin^2(t)} - 1 = 0[/math]or
[math]2 ~ cos(t) = 0[/math]

Here's a hint for @allegansveritatem concerning the case:-

[math]\dfrac{sin(t)}{1 - 2 ~ sin^2(t)} - 1 = 0[/math]
Put the LHS over a common denominator, and you'll end up with a quadratic in the numerator that can be factored by inspection.

 

[JeffM]

Here's a hint for @allegansveritatem concerning the case:-

[math]\dfrac{sin(t)}{1 - 2 ~ sin^2(t)} - 1 = 0[/math]
Put the LHS over a common denominator, and you'll end up with a quadratic in the numerator that can be factored by inspection.

Cubist

When I first looked at this problem back in post 2, what I thought was significant about it as a teaching tool was the utility of reducing a trigonometric equation to one involving a single trigonometric function with a common argument. That of course is the practical reason for teaching trigonometric identities. I was and remain confident of this OP's ability to solve this specific reduced equation (although I could not do it purely by inspection).

But look how wrong I was on what the problem eventually was able to teach this particular student. It turned out to be a much richer problem than I thought. My only fear now is that the utility of the initial reduction of a problem that did not appear to involve the sine function at all and involved two arguments to one involving only the sine function and a single argument may have got lost in the weeds.

 

[Cubist]

Cubist

When I first looked at this problem back in post 2, what I thought was significant about it as a teaching tool was the utility of reducing a trigonometric equation to one involving a single trigonometric function with a common argument. That of course is the practical reason for teaching trigonometric identities. I was and remain confident of this OP's ability to solve this specific reduced equation (although I could not do it purely by inspection).

But look how wrong I was on what the problem eventually was able to teach this particular student. It turned out to be a much richer problem than I thought. My only fear now is that the utility of the initial reduction of a problem that did not appear to involve the sine function at all and involved two arguments to one involving only the sine function and a single argument may have got lost in the weeds.

Stumbling into a pitfall is a great way of learning! And watching someone else getting caught out is equally educational ?. I think any of the helpers could have made such a mistake. Just slowly walk away before someone mentions the corner

 

[Steven G]

Here is the thing.

In my little stunt as you called it, I was ASSUMING IMPLICITLY that cos(t) was not zero because we cannot divide by zero in the real numbers. So if zero is a solution, performing that stunt will miss it. Therefore, before doing the stunt, you must test to see whether or not zero is a solution. Only after doing that can you use my stunt to look for non-zero solutions. Here is an example where it is obvious.

[MATH]x^6 - 16x^4 = 0 \implies x^2 - 16 = 0 \implies x = \pm 4.[/MATH]
THAT IS AN ERROR.

We missed the solution x = 0. The correct way is

[MATH]x = 0 \text { or } x \ne 0. \\ 0^6 - 16 * 0^4 = 0 - 0 = 0 \implies x = 0 \text { is a solution.}\\ x \ne 0 \text { and } x^6 - 16x^4 = 0 \implies \\ \dfrac{x^6 - 16x^4}{x^4} = \dfrac{0}{x^4} \implies x^2 - 16 = 0 \implies x = \pm 4. \\ \therefore x = - 4, \ x = 0, \text { or } x = 4.[/MATH]
So Dan was quite correctly pointing out in a very kind way that I was guilty of a major error.

I was going to let your error go, but if you are going to write that you were guilty of sloppy thinking (that is what I call it) then you must go to the corner for 1/0 minutes.