posted by Bryan .
Find all solutions to the following equation on the interval 0<=x<=2PI
8cos^2(X)sin^2(X) + 2cos^2(X) - 3 = 0
There are 8 solutions.
If somebody could show me how to do it and not give me the answers, that would be great.
Substitute sin^2(x) = 1-cos^2(x) and expand the expression in powers of
cos(x), You get a quadratic equation in cos^2(x). So, if you put cos^2(x) = y, the equation is quadratic in y.
First, solve the #1 Identity for sin^2(x) and plug it into the equation so that we have only cosines in the problem. The #1 Identity if sin^2(x) + cos^2(x) = 1.
Factor as you would if the cosine were simply x. Solve for cos(x).
Then, refer to the unit circle (which should be in your head) to solve for x. Cosine corresponds to the x-coordinate.
I only get 4 answers though. I'm supposed to get 8
If you put cos(x) = c, the equation becomes:
8c^4 - 10 c^2 + 3 = 0
(4 c^2 - 3)(2c^2 - 1) = 0
c = ±1/2 sqrt(3)
c = ±1/2 sqrt(2)
And you see that:
x = ±pi/6 + n pi
x = ±pi/4 + n pi
There are thus 8 solutions per interval of 2 pi.
ya sorry, I got it confused. Thanks for all the help.
if you have something like cos^2 x = 1/16
I am not picking what you actually have of course
then cos x = 1/4 OR cos x = -1/4
now cos 4 = 1/4 at 2 places on the unit circle (first quadrant and fourth)
cos x = -1/4 at two places on the circle (second quadrant and third)
that gives you four answers
The other solution for cos^2 gives you four more