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.
Distribute.
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)
and
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
To solve the equation 8cos^2(x)sin^2(x) + 2cos^2(x) - 3 = 0, we can use the fact that sin^2(x) = 1 - cos^2(x).
Let's substitute sin^2(x) with 1 - cos^2(x) in the given equation:
8cos^2(x)(1 - cos^2(x)) + 2cos^2(x) - 3 = 0
Now, simplify this equation:
8cos^2(x) - 8cos^4(x) + 2cos^2(x) - 3 = 0
Rearrange the equation:
-8cos^4(x) + 10cos^2(x) - 3 = 0
This can be thought of as a quadratic equation in terms of cos^2(x). If we let u = cos^2(x), then we have:
-8u^2 + 10u - 3 = 0
Now we can solve this quadratic equation. We can do this by factoring, completing the square, or using the quadratic formula. To factor or complete the square, we'll need to manipulate the equation to have a leading coefficient of 1. But since the quadratic formula works regardless, let's use it here.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In our case, a = -8, b = 10, and c = -3. Plug these values into the quadratic formula:
u = (-10 ± sqrt(10^2 - 4(-8)(-3))) / (2(-8))
Simplifying further:
u = (-10 ± sqrt(100 - 96)) / (-16)
u = (-10 ± sqrt(4)) / (-16)
We can simplify the square root:
u = (-10 ± 2) / (-16)
Now we have two possible solutions:
1) u = (-10 + 2) / (-16)
u = -8 / -16 = 1/2
2) u = (-10 - 2) / (-16)
u = -12 / -16 = 3/4
Since u = cos^2(x), we need to find the square root of these values to get cos(x):
1) cos(x) = sqrt(1/2) = √(1/2)
2) cos(x) = sqrt(3/4) = √(3/2)
Now, we consider the interval 0 <= x <= 2π.
For solution 1, cos(x) = √(1/2) has two possibilities: x = π/4 and x = 7π/4. These are valid solutions within the given interval.
For solution 2, cos(x) = √(3/2) is not possible within the range of -1 to 1. Therefore, there are no solutions for this case.
So, the solutions to the equation 8cos^2(x)sin^2(x) + 2cos^2(x) - 3 = 0 on the interval 0 <= x <= 2π are x = π/4 and x = 7π/4.
Keep in mind that there may be more solutions to this equation, but we focused on the given interval.