Solving For X In The Interval

Question

Solve for x in the interval...

 

0<x<2π: 3sin²(2x+1)+2sin(2x+1)-1=0

 

My teacher says there should be 6 answers.  I have a TI-84 calculator, so maybe there's a way to plug this in to find the solutions or make a graph of it?

Answer 1

This question is really really involved.

 

3*sin²(2x+1) + 2*sin(2x+1) - 1 = 0 looks kind of daunting. But you'll notice that sin(2x+1) appears in the first two terms, as a square and as itself.

 

Let's say A = sin(2x+1) and rewrite the equation.

 

3*A² + 2*A - 1 = 0

 

This looks a little more familiar and easy to manipulate.

 

(3A - 1)(A + 1) = 0

 

A = 1/3, -1

 

But we know from our substitution that A = sin(2x+1), so

 

sin(2x+1) = 1/3

 

sin(2x+1) = -1

 

Much better.

 

Your job is to find out where sin(2x+1) is equal to 1/3 and where it's equal to -1. Make sure you take into account the fact that you're taking sin(2x+1), and that 2 will make your period 2pi/2 = pi. Meaning in the given interval of length 2pi, sin(2x+1) will cycle twice.

 

Now that we've got that figured out:

 

sin(theta) = -1 at 3*pi/2

 

theta = 3pi/2

 

2x+1 = 3pi/2

 

2x = (3pi-2)/2

 

x = (3pi-2)/4

 

Since your period is pi rads, your next value will be

 

x = (3pi-2)/4 + pi = (7pi-2)/4

 

You can try to add pi again, but if you do that you'll be outside your interval.

 

(11pi-2)/4 > 2pi

 

 

So your two values from sin(theta) = -1 will be:

 

x = (3pi-2)/4

x = (7pi-2)/4

 

 

Now onto the next, much more painful one:

 

sin(theta) = 1/3 at arcsin(1/3), which is about pi/9 radians.

 

If you reference the unit circle or the graph of sine, you'll realize it will also occur at (pi - pi/9) radians, or, 8pi/9

 

Let's solve theta for pi/9 first:

 

theta = pi/9

 

2x+1 = pi/9

2x = pi/9 - 1

 

2x = (pi-9)/9

 

x = (pi-9)/18

 

Unfortunately, this is outside of our range of [0,2pi] for x. However, we can add a cycle to it to bring it back into range.

 

x = (pi-9)/18 + pi

 

   = (19pi-9)/18

 

Looks good. Let's add another cycle.

 

x = (19pi-9)/18 + pi

 

   = (37pi-9)/18

 

Looks good still. Let's try one more cycle.

 

x = (37pi-9)/18 + pi

 

   = (55pi-9)/18

 

This is > 2pi, so we're out of our range.

 

Let's move on to our other theta.

 

theta = 8pi/9

 

2x+1 = 8pi/9

 

2x = 8pi/9 - 1

 

2x = (8pi-9)/9

 

x = (8pi-9)/18

 

This looks good. Let's add another cycle:

 

x = (8pi-9)/18 + 18pi/18

 

   = (26pi-9)/18

 

Nice. Let's add another.

 

x = (26pi-9)/18 + 18pi/18

 

   = (44pi-9)/18

 

This is out of range as it's > 2pi

 

So let's put together our results:

 

From sin(2x+1) = -1

 

x = (3pi-2)/4
x = (7pi-2)/4

 

From sin(2x+1) = 1/3 [recall that pi/9 is an approximation]:

 

x = (19pi-9)/18

x = (37pi-9)/18

x = (8pi-9)/18

x = (26pi-9)/18

 

Those are your six answers, and it was pretty brutal. Thank goodness for graphing calculators to do this all for us.

 

What you'd want to do is plot your function in the window x:[0,2pi] and y:[-1,4] and see where it crosses the x-axis. There should be a function on your calculator called "zeros" which asks you to pick a value before and after the coordinate you're looking for. Do that for all 6 zeros and you'll get solutions that look exactly like the first two and pretty much like the last 4 above.

 

Hope this helps.