# SOlving y-intercept of a sin function

1. Sep 8, 2010

### t_n_p

1. The problem statement, all variables and given/known data

y = -2sin2(x+[pi/6])+1 for x [-pi, pi]

3. The attempt at a solution

set y=0,

1/2 = sin2(x+[pi/6])
pi/6 = 2(x+[pi/6]), where pi/6 is the base angle

Now because of the 2 infront of the (x+[pi/6]), I consider twice the domain, i.e. [-2pi, 2pi].
pi/6 is positive, and sin is positive in 1 & 2 quadrants.

therefore,
2(x+[pi/6]) = pi/6, 5pi/6, -7pi/6, -11pi/6.

divide by 2:
(x+[pi/6]) = pi/12, 5pi/12, -7pi/12, -11pi/12.

subtract [pi/6] from both sides:
x = -pi/12, 3pi/12, -9pi/12, -13pi/12.

This is my problem. Is -13pi/12 an issue since it is outside the original domain, or is it still ok since it is inside the modified domain [-2pi,pi]?

the answer has x values of -pi/12, 3pi/12, -9pi/12, 11pi/12, so my only issue is the last value. What has gone wrong!?

2. Sep 8, 2010

### CompuChip

-13pi/12 is a solution to the equation y = 0, but you are only asked to give the solutions between -pi and pi. However, if you add (or subtract) any multiple of 2pi to -13pi/12, you get an equivalent solution. In this case, you could find -13pi/12 + 2pi = 11pi/12, which is in the requested interval.

3. Sep 8, 2010

### t_n_p

ok, so I ask this, is there anyway to get 11pi/12 using the method I described, or can it only be obtained by adding 2pi?

so hypothetically, if I wanted x-intercepts for all x,
I would do:
-pi/12 +/- n*2pi
3pi/12 +/- n*2pi
-9pi/12 +/- n*2pi
-13pi/12 +/- n*2pi

where n is any integer?

I'm also wondering about the validity of adding pi (rather than 2pi). The period in our case is 2pi/2 = pi, so why doesn't adding/subtracing pi from all our values yield another number of solutions?

4. Sep 8, 2010

### CompuChip

The important thing in this type of problem, is to do everything in the right order.

Step 1: finding the base angle and writing down two solution sets
At the beginning of your post, you get to the above equation,
$$\sin 2(x + \pi/6) = \frac12 = \sin \pi/6$$
This equation has two "branches" of solutions, namely
$$2(x + \pi/6) = \pi / 6 + n \cdot 2\pi$$
and
$$2(x + \pi/6) = \pi - \pi / 6 + n \cdot 2\pi$$
for any integer n (= 0, 1, 2, 3, ..., -1, 2, -3, ...).
[If you draw the graph of the sine function or you look at its geometric meaning in a unit circle, you will see how the symmetry leads to the second equation].

You have to add the n*2pi here, because you want the sines of both sides to be equal, and if you add any multiple of 2 pi to any of them, this will be the case.

Step 2: solving for x
Now you can go and solve the two equations for x. The first one leads to
$$x + \pi/6 = \pi / 12 + n \cdot \pi$$
(note that nothing strange happens here, it is just basic algebra: if you divide everything by 2, then n*2pi changes into n*pi automatically. As you remarked, this corresponds to the period) and
$$x = \pi / 12 - \pi / 6 + n \cdot \pi = - \pi/12 + n \cdot \pi$$
For the second branch you will get something similar, which I will leave up to you to work out (be careful with all the minus signs though)

Step 3: Find the specific solutions requested
You now have two equations which describe infinitely many solutions, although basically there are just two and all the others differ from them by an integer number of 2pi steps.
In this case, you are specifically asked to list only the solutions between -pi and pi. So you can go and plug in some numbers for n into
$$x = - \pi/12 + n \cdot \pi$$
For n = 0 you get - pi / 12 which is in the interval, so you can write that down. For n = 1 you get - pi / 12 + pi = 11 pi / 12, that is also OK. For n = 2 you go above pi, and for n = -1 you are below -pi, so this is all the solutions you get.

Do the same for the other "branch" of solutions, and you will find two more.

5. Sep 8, 2010

### t_n_p

ok, that makes sense, but is this also feasible....

Follow what I did in my original post, where I found 4 solutions:

x = -pi/12, 3pi/12, -9pi/12, -13pi/12.

Why can't I just add multiples of the period to each of these values?

For x = -pi/12
x=-pi/12 + n*pi
test:
for n=-1, x=-13pi/12 (outside domain, invalid)
for n=0, x = -pi/12 (inside domain, valid)
for n=1, x= 11pi/12 (inside domain, valid)
for n=2, x= 23pi/12 (outside domain, invalid)

For x = 3pi/12
x=3pi/12 + n*pi
test:
for n=-2, x= -21pi/12 (outside domain, invalid)
for n=-1, x=-9pi/12 (inside domain, valid)
for n=0, x=3pi/12 (inside domain, valid)
for n=1, x= 15pi/12 (outside domain, invalid)

For x = -9pi/12
x=-9pi/12 + n*pi
test:
for n=-1, x=-21pi/12 (outside domain, invalid)
for n=0, x=-9pi/12 (inside domain, valid)
for n=1, x= 3pi/12 (inside domain, valid)
for n=2, x=15pi/12 (outside domain, invalid)

For x = -13pi/12
x=-13pi/12 + n*pi
test:
for n=-1, x=-25pi/12 (outside domain, invalid)
for n=0, x=-13pi/12 (outside domain, valid)
for n=1, x= -1pi/12 (inside domain, valid)
for n=2, x=11pi/12 (inside domain, valid)
for n=3, x=23pi/12 (outside domain, invalid)

then compiling all solutions that are valid I'm left with x=-9pi/12, 3pi/12, -pi/12, 11pi/12
as expected.

A little more time consuming but still works. My only question with the above method is, if the period is not pi, for example it is 2pi, will the same method work? I.e. if the period is pi, will adding n*2pi still provide the answers I am after?

6. Sep 8, 2010

### CompuChip

Yes, that also works. But where do you get these four solutions from in the first place? Are they guesses or what?

7. Sep 8, 2010

### t_n_p

I explained in the original post.

Now because of the 2 infront of the (x+[pi/6]), I consider twice the domain, i.e. [-2pi, 2pi]. I know the base angle is pi/6 and I know sin is positive in the 1st and 2nd quadrants. There will be 2 solutions per a revolution, then since there are 2 revolutions (domain is now -2pi to 2pi), there will be a total of 4 solutions.

That's the way I was taught. Since there is a factor of 2 out the front, I double the original domain. If there is a factor of 3 out the front, I triple the original domain so that it becomes -3pi to 3pi yielding a total of 6 solutions and so and so forth

8. Sep 8, 2010

### CompuChip

Ah, I see now. So to be entirely clear:

* If you already find four solutions this way, then you don't need to work everything out like in post #5. All you have to do is shift solutions outside the requested range by an integer number of periods (in this case, pi).

* Whenever you want to do the n * (2)pi thing: if you already divided everything by 2, you use multiples of pi. You only shift by 2pi, if you haven't divided by the 2 (or 3, or whatever number) in front yet.

9. Sep 8, 2010

### t_n_p

I definately prefer to do it the 1st star method.
Basically then, when I get a solution outside the bounds, I add or subtract 1 PERIOD (being careful to note what the period is, as it will change from problem to problem) to bring it back inside the bounds. E.g. if period is 500pi, then add/subtract 500pi from any values outside of the required bounds to bring it back inside.

This particular case had me stumped, since 3 of the solutions appear within the bounds, and only 1 is invalid. Obviously this is due to the value of the phase shift relative to the values obtained from the unit circle.

All makes sense now. Thanks