SOLUTION: Question: Solve over the interval [0 degrees, 360 degrees): 3cos^2x-4cosx-2=0 I have absolutely no idea how to do this problem. I tried using the quadratic formula but I kne

Algebra ->  Trigonometry-basics -> SOLUTION: Question: Solve over the interval [0 degrees, 360 degrees): 3cos^2x-4cosx-2=0 I have absolutely no idea how to do this problem. I tried using the quadratic formula but I kne      Log On


   

Question 817228: Question:
Solve over the interval [0 degrees, 360 degrees):
3cos^2x-4cosx-2=0
I have absolutely no idea how to do this problem. I tried using the quadratic formula but I knew that wasn't right once I did it that way. I then tried this way of doing it:
3cos^2x-4cosx-2=0
3cos^2x-4cosx=2
cosx(3cosx-4)=2
cosx=2/3cosx-4



Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
3cos%5E2%28x%29-4cos%28x%29-2=0+
You have a good idea with the factoring. But factoring an expression that is not equal to zero does us no good. So adding the 2 at the start was a mistake. Finding out that cos(x)(3cos(x)-4) = 2 does not help us at all.

The key is to factor the original equation (which has a zero on one side). It may not be easy to see how to factor this. It is in what is called quadratic from. This means the equation has the same structure as a regular quadratic, ax%5E2%2Bbx%2Bc=0, and can be solved using the same techniques.

To help you see the "quadratic-ness" of the equation, let's use a temporary variable:
Let q = cos(x). Replacing the cos(x)'s with q's we get:
3q%5E2-4q-2=0+
It should be obvious that we now have a quadratic equation. It does not factor but we can use the quadratic formula:
q+=+%28-%28-4%29%2B-sqrt%28%28-4%29%5E2-4%283%29%28-2%29%29%29%2F2%283%29
Simplifying...
q+=+%284%2B-sqrt%2816-4%283%29%28-2%29%29%29%2F6
q+=+%284%2B-sqrt%2816%2B24%29%29%2F6
q+=+%284%2B-sqrt%2840%29%29%2F6
q+=+%284%2B-sqrt%284%29%2Asqrt%2810%29%29%2F6
q+=+%284%2B-2%2Asqrt%2810%29%29%2F6
q+=+%282%282%2B-sqrt%2810%29%29%29%2F6
q+=+%282%2B-sqrt%2810%29%29%2F3
which is short for:
q+=+%282%2Bsqrt%2810%29%29%2F3 or q+=+%282-sqrt%2810%29%29%2F3

Of course we do not care about what q equals. We want to know what x equals. So we replace the q's:
cos%28x%29+=+%282%2Bsqrt%2810%29%29%2F3 or cos%28x%29+=+%282-sqrt%2810%29%29%2F3
The right sides of these equations are not special value numbers for cos. So we get out our calculators (and rounding to 2 places):
cos(x) = 1.7 or cos(x) = -0.39

Since cos is always between -1 and 1, the first equation is impossible. So we will not get any solutions from it. But the second equation is possible. First we find the reference angle. Using our calculators to find cos%5E-1%280.3874%29 we should get 67.05 for the reference angle. (Note that we did not use the "-". Never use minus signs when looking for a reference angle.) Since -0.3874 is negative (here is where the negative gets used) and since cos is negative in the 2nd and 3rd quadrants we should get the following general solution equations:
x+=+180-67.05%2B360n (for the 2nd quadrant)
x+=+180%2B67.05%2B360n (for the 3rd quadrant)
These simplify to:
x+=+112.95%2B360n
x+=+247.05%2B360n

Now we use these and various integers for n to find the x's that are in the specified interval.
From x+=+112.95%2B360n...
if n = 0 then x = 112.95
if n = 1 (or larger) then x is too large for the interval
if n = -1 (or smaller) then x is too small for the interval
From x+=+247.05%2B360n...
if n = 0 then x = 247.05
if n = 1 (or larger) then x is too large for the interval
if n = -1 (or smaller) then x is too small for the interval

So the only solutions that are in the specified interval are 112.95 and 247.05 degrees.

P.S. After you have done several of these quadratic form equations you will no longer need the temporary variable. You will start seeing how to go directly from:
3cos%5E2%28x%29-4cos%28x%29-2=0+
to
cos%28x%29+=+%28-%28-4%29%2B-sqrt%28%28-4%29%5E2-4%283%29%28-2%29%29%29%2F2%283%29
etc.