# Thread: Help with solving a trigonometric equation

1. ## Help with solving a trigonometric equation

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

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

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

x ints, y=0

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

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

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

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

Dividing through by 3

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

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

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

2. Hello user_5
Originally Posted by user_5
Alright, so I am trying to solve the function for x:

$\displaystyle y=2\sin(3x) - 2 x \in[-2\pi,2\pi]$
I don't know what you mean by 'solve'. This is $\displaystyle y$ written as a function of $\displaystyle x$.

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

x ints, y=0

$\displaystyle 0=2\sin(3x) -2$
$\displaystyle 1=\sin(3x)$
I don't understand this either. You have put $\displaystyle y = 0$, but what about the term in $\displaystyle 2x$? This just appears as $\displaystyle 2$ now.

So I don't think any of this:
Basic angle $\displaystyle = \sin^{-1}(1) = \frac{\pi}{2}$

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

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

Dividing through by 3

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

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

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

Could we have the original question again please?

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

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 $\displaystyle x$ the subject of $\displaystyle y = 2\sin3x -2x$.

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

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

6. 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 $\displaystyle x$ between $\displaystyle -2\pi$ and $\displaystyle 0$ that satisfy $\displaystyle \sin3x = 1$.

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

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

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

7. Thanks, but how did you get those answers?
$\displaystyle x=-\frac{\pi}{2},-\frac{7\pi}{6},-\frac{11\pi}{6}$

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 $\displaystyle x$ value
2nd quadrant = $\displaystyle \pi - x$
3rd quadrant = $\displaystyle \pi+ x$
4th quadrant = $\displaystyle 2\pi- x$

8. Hello user_5
Originally Posted by user_5
Thanks, but how did you get those answers?
$\displaystyle x=-\frac{\pi}{2},-\frac{7\pi}{6},-\frac{11\pi}{6}$

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 $\displaystyle x$ value
2nd quadrant = $\displaystyle \pi - x$
3rd quadrant = $\displaystyle \pi+ x$
4th quadrant = $\displaystyle 2\pi- x$
Using degrees rather than radians (it's easier!) start at $\displaystyle 0^o$ at the point $\displaystyle (1,0)$ on the unit circle. Rotate $\displaystyle 90^o$ clockwise to the point $\displaystyle (0,-1)$: $\displaystyle y$-value $\displaystyle = -1$. So $\displaystyle \sin(-90^o) = -1$.

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

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

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

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