## Find Solutions 2 Parts

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

### Find Solutions 2 Parts

I've been having some problems with my most recent section, I'm supposed to find all of the solutions to the following functions and be able to name all the solutions within [0, 2(pi)].

$cos\,(2x\,-\,((pi)/4))\,=\,0$
The parenthesis are throwing me off; should I take the cos^-1? To be left with (2x-((pi)/4)) and solve from there? That gets messy though.
and
$12sin^2\,u\,-\,5sinu\,-\,2\,=\,0$
This looks like a quadratic, I tried factoring out 5sinu and it didn't end well, but I don't know how to deal with it as a quadratic.

Those seem to be the two distinct ways they are written in my book.

My book is showing a new equation that should graph a line through all of the solutions as the answer. After I get the new function, the [0, 2(pi)] should be as simple as plugging in values right? My only concern is that I need to plug in values other than integers.
GreenLantern

Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm

. . . . .$\cos(\theta)\, =\, 0$

...what would have been your solution? (Since you're actually dealing with a different argument, find all the solutions in, say, $-\pi$ to $3\pi$.)

Once you have your list of solutions, set the given argument equal to the list of values:

. . . . .$2x\, -\, \frac{\pi}{4}\, =\, \theta$

Solve for $x$.

stapel_eliz

Posts: 1797
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Find Solutions 2 Parts

Taking that, I have:
$\frac{\pi}{2}\,\,\, \frac{3\pi}{2}\,\,\, \frac{5\pi}{2}\,\,\, \frac{-\pi}{2}\$
Solve for each, so now I have:
$\frac{3\pi}{8}\,\,\, \frac{7\pi}{8}\,\,\, \frac{11\pi}{8}\,\,\, \frac{-\pi}{8}\$
for my answers; my book is giving me:
$\frac{15\pi}{8}\$
as the last answer, but they are really the same thing.

The equation I'm supposed to come up with is:
$\frac{3\pi}{8}\,+\,\frac{4\pi}{8}\,n\,=\,\theta$
It hits all of the possible solutions for my original equation. I came up with it by finding the difference between each of the 4 answers (simple subtraction); I don't feel like that's how I was supposed to come up with it, is that just supposed to work for all of them?

Now the second equation I posted. I've been doing some work with it since I posted and have gotten to:
$12sin^2\,u\,-\,5sinu\,-\,2\,=\,0$
I replaced sinu with z to get:
$12z^2\,-\,5z\,-\,2\,=\,0$
$\frac{+\,5\,\pm\,i\,\sqrt{\,96}}{24}$
But I don't know what that gets me. If I'm plugging that into u it won't equal 0 anymore, and I get a big ol' error on my calculator.
GreenLantern

Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm

GreenLantern wrote:Taking that, I have:
$\frac{\pi}{2}\,\,\, \frac{3\pi}{2}\,\,\, \frac{5\pi}{2}\,\,\, \frac{-\pi}{2}\$
Solve for each, so now I have:
$\frac{3\pi}{8}\,\,\, \frac{7\pi}{8}\,\,\, \frac{11\pi}{8}\,\,\, \frac{-\pi}{8}\$
for my answers; my book is giving me:
$\frac{15\pi}{8}\$
as the last answer, but they are really the same thing.

The fact that you got the same thing but for the last answer, which was "past" where you'd checked, indicates only that we didn't expand our original interval sufficiently. Since you ended up dividing by 2, we probably should have expanded to $4\pi$ rather than just to $3\pi$. In other words, the method was right, but we didn't correct sufficiently.

GreenLantern wrote:Now the second equation I posted. I've been doing some work with it since I posted and have gotten to:
$12sin^2\,u\,-\,5sinu\,-\,2\,=\,0$

Since (12)(-2) = -24 can be factored as (-8)(+3), which (-8) + (+3) = -5, this does factor. Then you can solve the trig equations formed by setting the factors equal to zero.

stapel_eliz

Posts: 1797
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Find Solutions 2 Parts

Okay, maybe I'm not to great at recognizing when I can factor; I'll be putting some more practice into that I guess. After factoring I should have:
$(sinu\,-\,8)\,(sinu\,+\,3)\,=\,0$
Set the factors equal to zero...
$(sinu\,-\,8)\,=\,0$
$(sinu\,+\,3)\,=\,0$
Solve...
$sinu\,=\,8$
$sinu\,=\,-\,3$
Both of those are not within [-1,1], so sin cannot equal either with any variables; taking the inverse sin gets the same results. Normally I'd just say it has no answer because sinu cannot equal anything outside the bounds [-1,1], but this one has real answers in the back of my book. I really am stumped by this one. I've read over the chapter again and there is no mention of dealing with this sort of thing (quadratics or -1>sinu>1). I've been trying to go backwards by taking my book answer and just plugging it in, but it's just some decimal (neither 3 nor 8).

Okay, new idea. I went to far, go back to
$(sinu\,-\,8)\,=\,0$
$(sinu\,+\,3)\,=\,0$
$sinu\,=\,0$
$u\,=\,\frac{\pi}{2}\,n$
That means...
$0$,$\pi$,$2\pi$...
Now add -8 and 3 respectively?
$u\,=\,\frac{\pi}{2}\,n\,-\,8$
$u\,=\,\frac{\pi}{2}\,n\,+\,3$
But (I'm only guessing) based on what you gave me before
stapel_eliz wrote:$2x\, -\, \frac{\pi}{4}\, =\, \theta$

My n variable should be on the -8 and 3, right?
$u\,=\,\frac{\pi}{2}\,-\,8\,n$
$u\,=\,\frac{\pi}{2}\,+\,3\,n$
So I've got some very slow progress, but I have
$\frac{\pi}{16}$,$\frac{-\,\pi}{16}$,$\frac{-\,3\pi}{16}$
$\frac{-\,\pi}{6}$,$\frac{\pi}{6}$,$\frac{3\pi}{6}$
But none of the answers are the ones in the book.
GreenLantern

Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm

GreenLantern wrote:Okay, maybe I'm not to great at recognizing when I can factor; I'll be putting some more practice into that I guess. After factoring I should have:
$(sinu\,-\,8)\,(sinu\,+\,3)\,=\,0$

You seem to have dropped the leading "12"...?

To learn how to factor quadratics, try this lesson. Your equation is covered in "The Hard Case", but you'll want to learn "The Easy Case" first, so that "The Hard Case" makes sense.

Once you have studied that lesson, try the factorization again. You have (12)(-2) = -24, with factors -8 and +5 which add to -3.

GreenLantern wrote:My n variable should be on the -8 and 3, right?

I'm not clear on what you're doing here, so I'm afraid I can't comment intelligently. Sorry!

stapel_eliz

Posts: 1797
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Find Solutions 2 Parts

Okay, I finally got through this thing, and just in time; I had the test on it earlier today! The link to the box? Amazing! I had no idea.

Alright, down to business. From the top.
$12sin^2\,u\,-\,5sinu\,-\,2\,=\,0$
"Box it", probably my favorite term for this chapter.
$4x\,+\,1\,=\,0$
$3x\,-\,2\,=\,0$
Solve
$Sinu\,=\,\frac{2}{3}$
$Sinu\,=\,\frac{-\,1}{4}$
Invert that function
$u\,=\,sin^{-1}\,\frac{2}{3}$
$u\,=\,sin^{-1}\,\frac{-\,1}{4}$
That equals seemingly random decimals, and I was stuck here for awhile until I realized they aren't random. They are in radians right now! I totally didn't recognize them, so I changed those into degrees to get:
$41.8103$
$-14.4775$
(Both rounded to four decimal places.) But these are only two of the four answers I need. Well, a friend of mine (in the same class) told me "sin=y cos=x according to the axis" that's all that matters. What that means is, sin is positive in quadrants one and two (just like y), while cos is positive in one and four (just like x). The first answer I have is positive according to sin (which puts it above the y axis) so one of my missing answers is also positive (and therefore also above the y axis), but in the other quadrant. To find it I need to subtract my first answer from 180.
$180\,-\,41.8103\,=\,138.1897$
On a unit circle it looks like I'm just going backwards 41.8103 from 180 (all in degrees by the way).
My last missing answer is done in much the same way as the first, but I need to put it in the fourth quadrant. Simply take the absolute value of the second answer and add it to 180.
$14.4775\,+\,180\,=\,194.4775$
That makes all four of my answers:
$41.8103\,=\,41^o\,50'$
$138.1897\,=\,138^o\,10'$
$194.4775\,=\,194^o\,30'$
$-14.4775\,=\,360^o\,-\,14.4775\,=345^o\,50'$
All exactly the same as my book answers!

I really don't know if anyone will understand that middle paragraph, but I get it! It has worked for all the other ones in my book too.

Anyway, thanks Liz. When I started I was waaay off on everything, but you pushed me in all the right directions. Thanks again!
GreenLantern

Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm