Solving trig equations involving aperiodic functions

[Serious Max]

Homework Statement

Homework Equations

—

The Attempt at a Solution

C)

Principal domain:

[tex]-\dfrac{\pi}{2}\leq 2x^2+x-1\leq \dfrac{\pi}{2}[/tex]

[tex]-1.41099 \leq x \leq 0.91099[/tex]

Next:

[tex]\sin(2x^2+x-1)=\dfrac{2}{5}[/tex]

[tex]2x^2+x-1=\arcsin(\dfrac{2}{5})[/tex]

[tex]x_1=-1.1265; x_2=0.62650[/tex]

Since [tex]\sin(\theta)=-\sin(\theta)+\pi[/tex] we get:

[tex]\sin(-(2x^2+x-1)+\pi)=\dfrac{2}{5}[/tex]

[tex]-2x^2-x+1+\pi=\arcsin(\dfrac{2}{5})[/tex]

Hence:

[tex]x_3=1.1384; x_4=-1.6384[/tex]

Although, I wouldn't be able to find any more values of x by the methods I have been introduced to, since the functions is aperiodic.By the same method I can find two values of x in D), but that's it. Although, I think they want me to do something else, rather than values for x.
P.S. Also, I haven't been introduced to solving trig equations with aperiodic functions, so it must be that they want to stretch me a little. So the tasks are probably rigged somehow, like in the case of C) I can find exactly four values with the methods I've been shown (well, as I understand), and that's what they want from me in the problem.

So, I guess it must be similar with the D) task, but I can't figure it out.

 

[micromass]

If ##\sin(a) = \frac{2}{5}##, what are all the possible values for ##a##?

 

[micromass]

Principal domain:

[tex]-\dfrac{\pi}{2}\leq 2x^2+x-1\leq \dfrac{\pi}{2}[/tex]

Where does this come from? Why must ##x## satisfy this? This isn't given in the question.

 

[Serious Max]

It's just a procedure from my book to find 1 value (well, in this case there are two, because this problem involves a quadratic function) on this restricted domain and then work from it deriving general solutions.

If ##\sin(a) = \frac{2}{5}##, what are all the possible values for ##a##?

Well, in this case
## a=-(\arcsin(\dfrac{2}{5})-\pi) ##

 

[Serious Max]

Oh wait, so I guess then:

## 2x^2+x-1=-(\arcsin(\dfrac{2}{5})-\pi) ## ?

 

[micromass]

It's just a procedure from my book to find 1 value (well, in this case there are two, because this problem involves a quadratic function) on this restricted domain and then work from it deriving general solutions.



Well, in this case
## a=-(\arcsin(\dfrac{2}{5})-\pi) ##

Those are not all the possible values for ##a##, but ok, setting

[tex]2x^2 + x -1 = -(\arcsin(\dfrac{2}{5})-\pi)[/tex]

should work.

 

[Serious Max]

Yes, I fixed it, forgot to include something:

## a=-(\arcsin(\dfrac{2}{5})-\pi +2k \pi), k \in \mathbb Z ##

 

[micromass]

Yes, I fixed it, forgot to include something:

## a=-(\arcsin(\dfrac{2}{5})-\pi +2k \pi), k \in \mathbb Z ##

Those are still not all the solutions, for example ##\arcsin(2/5)## isn't included in there.

 

[Serious Max]

Right... Silly me. But I get the idea.

## a=\arcsin(\dfrac{2}{5})+2k \pi, k \in \mathbb Z #### a=-\arcsin(\dfrac{2}{5})+\pi +2k \pi, k \in \mathbb Z ##

 

[micromass]

Right... Silly me. But I get the idea.

## a=\arcsin(\dfrac{2}{5})+2k \pi, k \in \mathbb Z ##


## a=-\arcsin(\dfrac{2}{5})+\pi +2k \pi, k \in \mathbb Z ##

Right, and all individual values for k give us a solution. So this can be used to find many solutions of your original equation.

 

[Serious Max]

Oh, actually I just noted that ## k \notin \mathbb Z## but rather ## k \in \mathbb W ##. Otherwise the solutions are complex numbers. And any ##k \in \mathbb W## will give both negative and positive solutions.

Has to do with the "quadraticity" of the function.

 

[SammyS]

Oh, actually I just noted that ## k \notin \mathbb Z## but rather ## k \in \mathbb W ##. Otherwise the solutions are complex numbers. And any ##k \in \mathbb W## will give both negative and positive solutions.

Has to do with the "quadraticity" of the function.

What is ##\mathbb W \ ?##

 

[Serious Max]

Whole numbers. I wasn't sure if you could denote it like that, but I did :).

Well, I guess it actually isn't allowed, so then it should be ## k \in \mathbb N, k=0 ##

Or maybe like this http://mathworld.wolfram.com/Z-Star.html

So ## k \in \mathbb{Z}^+ ##