Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. (Enter your answers as a comma-separated list.)

cos(x)(9cos(x) + 4) = 4; [0, 2π)

expand first

9cos^2 x + 4cosx - 4 = 0
let y = cosx
9y^2 + 4y - 4 = 0
y = (-4 ±√160)/18 , using the formula
= (-4 + 4√10)/18
= (-2 ± 2√10)/9
= appr .480506 or appr -.92495

then cosx = .480506 or cosx = -.92495
x = 61.28° or 298.72° or 157.66° or 202.34°

To solve the given equation, we need to use inverse trigonometric functions.

Let's start by simplifying the equation:
cos(x)(9cos(x) + 4) = 4

Expanding the equation, we get:
9cos^2(x) + 4cos(x) - 4 = 0

Now, let's use a substitution to make the equation simpler.
Let's set cos(x) = t. So, the equation becomes:
9t^2 + 4t - 4 = 0

Next, we can solve this quadratic equation for t. Use the quadratic formula to find the value of t:
t = (-b ± √(b^2 - 4ac))/(2a)

For our equation:
a = 9, b = 4, c = -4

Plugging these values into the quadratic formula, we have:
t = (-4 ± √(4^2 - 4*9*(-4))) / (2*9)
t = (-4 ± √(16 + 144)) / 18
t = (-4 ± √160) / 18
t = (-4 ± 4√10) / 18
t = (-2 ± 2√10) / 9

Now, remember that we chose t = cos(x). So, we can equate cos(x) to these values and find the solutions. To find the value of x, we need to use the inverse trigonometric function, arccos.

Using the arccos function, we have:
x = arccos((-2 ± 2√10) / 9)

Now, we need to find x within the given interval of [0, 2π). To do this, we need to find the values of x that satisfy the equation and are within the interval.

Let's calculate the approximate values of x using a calculator or a math software:

x ≈ arccos((-2 + 2√10) / 9) ≈ 0.8609
x ≈ arccos((-2 - 2√10) / 9) ≈ 2.2819

Since these values are within the interval [0, 2π), these are the solutions within the given interval.

Therefore, the approximate solutions to the equation cos(x)(9cos(x) + 4) = 4 within the interval [0, 2π) are approximately 0.8609 and 2.2819.