find the solutions of the equation that are in the interval [0,2pi)

2-cos^2x=4sin^2(1/2x)

2 - (1 - 2sin^2 (x/2) = 4sin^2 (x/2)

1 = 2 sin^2 (x/2)
sin^2 (x/2) = 1/2
sin (x/2) = 1/√2
x/2 = π/4 or x/2 = 3π/4

then x = π/2 or x = 3π/2

To solve the given equation, follow these steps:

Step 1: Simplify the equation using trigonometric identities.
2 - cos²x = 4sin²(1/2x)

Using the double angle formula for sin(2θ), we have:
2 - cos²x = 4(1 - cos²(1/2x))

Now, substitute cos²(1/2x) with (1 + cos(x)) / 2:
2 - cos²x = 4(1 - (1 + cos(x)) / 2)

Simplifying this further gives:
2 - cos²x = 4/2 - 4cos(x)/2
2 - cos²x = 2 - 2cos(x)

Step 2: Rearrange the equation and combine like terms.
cos²x - 2cos(x) + 2 = 0

Step 3: Solve the quadratic equation.
To solve this quadratic equation, let's use the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)

In our case, a = 1, b = -2, and c = 2. Substituting these values into the quadratic formula, we have:
x = (-(-2) ± √((-2)² - 4(1)(2))) / (2(1))
x = (2 ± √(4 - 8)) / 2
x = (2 ± √(-4)) / 2

Step 4: Determine the solutions.
Since we have a negative value under the square root, the equation has no real solutions. In the given interval of [0, 2π), there are no values of x that satisfy the equation.