Help with solving a trigonometric equation

**user_5** · October 14th 2009, 03:56 AM

I am trying to find the x-intercepts of this function.

$y=2\sin(3x) - 2 \in[-2\pi,2\pi]$

Okay, so I make the domain $3x \in[-6\pi,6\pi]$

x ints, y=0

$0=2\sin(3x) -2$
$1=\sin(3x)$

Basic angle $= \sin^{-1}(1) = \frac{\pi}{2}$

$\sin$ is positive in 1st + 2nd quadrants.

$3x=\sin^{-1}(1)$
$3x=\frac{\pi}{2}$ (repeated) , $\frac{5\pi}{2}$ (repeated) , $\frac{9\pi}{2}$ (repeated)

Dividing through by 3

$x=\frac{\pi}{6}$ (repeated) , $\frac{5\pi}{6}$ (repeated) , $\frac{3\pi}{2}$ (repeated)

So these are the intercepts from $[0,2\pi]$

My question is, how do I get the intercepts from $[-2\pi,0]$ ?

**Grandad** · October 14th 2009, 05:41 AM

Hello user_5

Originally Posted by user_5

Alright, so I am trying to solve the function for x:

$y=2\sin(3x) - 2 x \in[-2\pi,2\pi]$

I don't know what you mean by 'solve'. This is $y$ written as a function of $x$ .

Okay, so I make the domain $3x \in[-6\pi,6\pi]$

x ints, y=0

$0=2\sin(3x) -2$
$1=\sin(3x)$

I don't understand this either. You have put $y = 0$ , but what about the term in $2x$ ? This just appears as $2$ now.

So I don't think any of this:

Basic angle $= \sin^{-1}(1) = \frac{\pi}{2}$

$\sin$ is positive in 1st + 2nd quadrants.

$3x=\sin^{-1}(1)$
$3x=\frac{\pi}{2}$ (repeated) , $\frac{5\pi}{2}$ (repeated) , $\frac{9\pi}{2}$ (repeated)

Dividing through by 3

$x=\frac{\pi}{6}$ (repeated) , $\frac{5\pi}{6}$ (repeated) , $\frac{3\pi}{2}$ (repeated)

So these are the intercepts from $[0,2\pi]$

My question is, how do I get the intercepts from $[-2\pi,0]$ ?

makes much sense.

Could we have the original question again please?

Grandad

**adkinsjr** · October 14th 2009, 07:10 AM

Originally Posted by Grandad

Hello user_5I don't know what you mean by 'solve'. This is $y$ written as a function of $x$ .

Well, he said that he wants to solve the function for x. So I think he is trying to find the inverse of y. But I don't understand what he wrote after that either.

Actually, I think that would be impossible. You can't isolate x.

**Grandad** · October 14th 2009, 07:16 AM

Hello adkinsjr

Originally Posted by adkinsjr

Well, he said that he wants to solve the function for x. So I think he is trying to find the inverse of y. But I don't understand what he wrote after that either.

Actually, I think that would be impossible. You can't isolate x.

Agreed. You cannot make $x$ the subject of $y = 2\sin3x -2x$ .

That's why I don't understand what 'solve' means in this context.

Grandad

**user_5** · October 14th 2009, 09:05 PM

Sorry, I fixed the typo it is just -2 instead of -2x.

**Grandad** · October 14th 2009, 10:24 PM

Hello user_5

Originally Posted by user_5

Sorry, I fixed the typo it is just -2 instead of -2x.

OK, so you want to know the values of $x$ between $-2\pi$ and $0$ that satisfy $\sin3x = 1$ .

You have already noted that if $-2\pi \le x \le 0$ , then $-6\pi \le3x \le0$ , so the values of $3x$ are:

$3x=-\frac{3\pi}{2},-\frac{7\pi}{2}, -\frac{11\pi}{2}$

$\Rightarrow x=-\frac{\pi}{2},-\frac{7\pi}{6},-\frac{11\pi}{6}$

Grandad

**user_5** · October 14th 2009, 10:51 PM

Thanks, but how did you get those answers?
$<br /> x=-\frac{\pi}{2},-\frac{7\pi}{6},-\frac{11\pi}{6}<br />$

I'm not exactly sure how to go in the clockwise direction of the unit circle.

Like if you were going anti clockwise, the answer in the
1st quadrant = Just the $x$ value
2nd quadrant = $\pi - x$
3rd quadrant = $\pi+ x$
4th quadrant = $2\pi- x$

**Grandad** · October 14th 2009, 11:36 PM

Hello user_5

Originally Posted by user_5

Thanks, but how did you get those answers?
$<br /> x=-\frac{\pi}{2},-\frac{7\pi}{6},-\frac{11\pi}{6}<br />$

I'm not exactly sure how to go in the clockwise direction of the unit circle.

Like if you were going anti clockwise, the answer in the
1st quadrant = Just the $x$ value
2nd quadrant = $\pi - x$
3rd quadrant = $\pi+ x$
4th quadrant = $2\pi- x$

Using degrees rather than radians (it's easier!) start at $0^o$ at the point $(1,0)$ on the unit circle. Rotate $90^o$ clockwise to the point $(0,-1)$ : $y$ -value $= -1$ . So $\sin(-90^o) = -1$ .

Rotate a further $90^o$ to $-180^o: \sin(-180^o)=0$

A further $90^o$ to $-270^o: \sin(-270^o) = 1$ . That's the first one we want!

Every $360^o$ further takes you to the same point: $(0,1)$ .

In radians, then, all the following have a sine of $1: -\frac{3\pi}{2},\,-\frac{3\pi}{2}-2\pi=-\frac{7\pi}{2},\,-\frac{7\pi}{2}-2\pi=-\frac{11\pi}{2}, ...$

Grandad

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