Solving trig equations involving aperiodic functions

[Serious Max]

Homework Statement

Homework Equations

—

The Attempt at a Solution

C)

Principal domain:

$$-\dfrac{\pi}{2}\leq 2x^2+x-1\leq \dfrac{\pi}{2}$$

$$-1.41099 \leq x \leq 0.91099$$

Next:

$$\sin(2x^2+x-1)=\dfrac{2}{5}$$

$$2x^2+x-1=\arcsin(\dfrac{2}{5})$$

$$x_1=-1.1265; x_2=0.62650$$

Since $$\sin(\theta)=-\sin(\theta)+\pi$$ we get:

$$\sin(-(2x^2+x-1)+\pi)=\dfrac{2}{5}$$

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

Hence:

$$x_3=1.1384; x_4=-1.6384$$

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:

$$-\dfrac{\pi}{2}\leq 2x^2+x-1\leq \dfrac{\pi}{2}$$

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

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

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 $k \in \mathbb{Z}^+$