Hi, can someone please solve this step-by-step for me? I really don't get it...

Directions: Use the quadratic formula to solve the equation in the interval [0,2PI). Then use a graphing utility to approximate the angle x.

4cos^2x-4cosx-1=0

thanks.

LET Y = COS X

Then you have

4y^2-4y-1=0

Then use the quadratic equation to solve for y. then put for y, the original value...cosx= solution, and solve for x.

Sure, I can help you with that. To solve the equation 4cos^2x - 4cosx - 1 = 0, we can use the quadratic formula, which is:

x = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 4, b = -4, and c = -1. Let's plug these values into the formula and solve for x step by step.

1. Calculate the discriminant: b^2 - 4ac
Discriminant = (-4)^2 - 4(4)(-1) = 16 + 16 = 32

2. If the discriminant is positive, there are two real solutions. If it is zero, there is one real solution. If it is negative, there are no real solutions. Since the discriminant is positive (32 > 0), we will have two real solutions.

3. Use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
x = (-(-4) ± √(16 - 4(4)(-1))) / (2(4))
= (4 ± √(16 + 16)) / 8
= (4 ± √(32)) / 8

4. Simplify the square root: √(32) = √(16 * 2) = 4√2

5. Rewrite the equation: x = (4 ± 4√2) / 8

6. Simplify the expression: x = (1 ± √2) / 2

So the two real solutions for x, in the interval [0, 2π), are:

x1 = (1 + √2) / 2
x2 = (1 - √2) / 2

To approximate the angles, you can use a graphing utility like a calculator or software that can plot the cosine function. Plot the graph of y = 4cos^2x - 4cosx - 1, and find the x-values where the graph intersects the x-axis between 0 and 2π. These points will give you the approximate angles where the equation is satisfied.