Solve for [0,2pi)

2cos4x + sqrt3 = 0

2cos4x + √3 = 0

cos4x = -√3/2
4x = 5π/6 or 7π/6
x = 5π/24 or 7π/24
period is 2π/4 = π/2, so all solutions are

5π/24 + kπ/2 or 7π/24 + kπ/2

So, for all solutions in [0,2π) we have

5π/24, 17π/24, 29π/24, 41π/24
7π/24, 19π/24, 31π/24, 43π/24

To solve the equation 2cos(4x) + √3 = 0 over the interval [0, 2π), we will follow these steps:

Step 1: Isolate the cosine term
First, let's isolate the cosine term by subtracting √3 from both sides of the equation:
2cos(4x) = -√3

Step 2: Divide by 2
Next, divide both sides of the equation by 2 to solve for cos(4x):
cos(4x) = -√3/2

Step 3: Find the angle(s)
Now, we need to find the angles within the interval [0, 2π) that have a cosine value of -√3/2. We can refer to the unit circle or the special trigonometric ratios to find these angles.

On the unit circle, we know that cos(θ) = x-coordinate. So, we are looking for angles where the x-coordinate is -√3/2.

The reference angle for -√3/2 is π/6. In the unit circle, this corresponds to the angle π/6.

Step 4: Determine the solutions
Since cosine is a periodic function with a period of 2π, we need to find all possible solutions within the interval [0, 2π). The cosine function is positive in Quadrant I and Quadrant IV.

In Quadrant I, the angle with a cosine value of -√3/2 is π/6.
In Quadrant IV, the angle with a cosine value of -√3/2, using the reference angle π/6, is 11π/6.

So, the two solutions within the interval [0, 2π) are π/6 and 11π/6.

Therefore, the solutions to the equation 2cos(4x) + √3 = 0 over the interval [0, 2π) are x = π/24 and x = 11π/24.

To solve the equation 2cos(4x) + √3 = 0 in the interval [0, 2π), follow these steps:

Step 1: Subtract √3 from both sides of the equation:
2cos(4x) = -√3

Step 2: Divide both sides of the equation by 2:
cos(4x) = -√3/2

Step 3: Find the inverse cosine (arccos) of both sides:
4x = arccos(-√3/2)

Step 4: Solve for x by dividing both sides by 4:
x = arccos(-√3/2) / 4

Step 5: Use a calculator to find the value of arccos(-√3/2) and divide it by 4. The principal value of arccos(-√3/2) is π/6.

Step 6: Divide π/6 by 4 to get the solution for x:
x = π/6 / 4 = π/24

So, the solution to the equation 2cos(4x) + √3 = 0 in the interval [0, 2π) is x = π/24.