Solve the equation. (Find all solutions of the equation in the interval [0, 2π). Enter your answers as a comma-separated list.)

4 tan(2x) − 4 cot(x) = 0

I put that in and it said it wasn't correct. Is there anything that could be missing?

well, π/6 works fine, since

4(√3 - √3) = 0

To solve the equation 4 tan(2x) - 4 cot(x) = 0, we can use trigonometric identities to express tan(2x) and cot(x) in terms of sine and cosine.

First, let's rewrite tan(2x) and cot(x) using the basic trigonometric identities:
tan(2x) = sin(2x) / cos(2x)
cot(x) = 1 / tan(x) = cos(x) / sin(x)

Substituting these expressions into our equation gives:
4 * (sin(2x) / cos(2x)) - 4 * (cos(x) / sin(x)) = 0

Now, let's simplify this equation by getting rid of the fractions. Multiply both sides of the equation by the least common denominator, which is sin(x) * cos(2x):

4 * sin(2x) * sin(x) - 4 * cos(x) * cos(2x) = 0

Expanding the terms gives:
4 * (2 * sin(x) * cos(x)) * sin(x) - 4 * cos(x) * (2 * cos^2(x) - 1) = 0

Simplifying further:
8 * sin^2(x) * cos(x) - 8 * cos(x) * (2 * cos^2(x) - 1) = 0

Combining like terms:
8 * sin^2(x) * cos(x) - 16 * cos^3(x) + 8 * cos(x) = 0

Now, let's factor out a common term:
8 * cos(x) * (sin^2(x) - 2 * cos^2(x) + 1) = 0

The equation is now factored, and we can set each factor equal to zero:

cos(x) = 0 (equation 1)
sin^2(x) - 2 * cos^2(x) + 1 = 0 (equation 2)

Starting with equation 1, we know that cos(x) = 0 has solutions when x = π/2 and x = 3π/2 in the interval [0, 2π).

Now, let's solve equation 2. We'll make a substitution using the Pythagorean identity sin^2(x) + cos^2(x) = 1:

1 - 2 * cos^2(x) + 1 = 0

Rearranging terms gives:
2 * cos^2(x) = 2

Dividing both sides by 2, we have:
cos^2(x) = 1

Taking the square root of both sides and considering the positive and negative solutions gives:
cos(x) = 1 or cos(x) = -1

From this, we conclude that x = 0, x = π, and x = 2π are solutions for cos(x) = 1, and x = π/2 and x = 3π/2 are solutions for cos(x) = -1 in the interval [0, 2π).

Combining all the solutions, we have:
x = 0, x = π/2, x = π, x = 3π/2, and x = 2π.

Therefore, the solutions of the given equation in the interval [0, 2π) are:
0, π/2, π, 3π/2, and 2π.

tan(2x) = 2tanx/(1-tan^2 x)

let tanx = k , then

4 tan(2x) − 4 cot(x) = 0
4(2k)/(1 - k^2) - 4/k = 0
8k/(1 - k^2) = 4/k
2k(1 - k^2) = 1/k
2k^2 = 1 - k^2
3k^2 = 1
k = ±1/√3

tanx = ±1/√3
so x could be in any of the 4 quadrants,
x = 30°, 150° , 210°, 330°
or
x = π/6, 5π/6, 7π/6, 11π/6