Thank you for the clarification. The equation to solve is:

\[
2\cos\left(2\left(x + \frac{\pi}{3}\right)\right) = -1
\]

Let's solve this equation step by step.

### Step 1: Simplify the equation
First, divide both sides by 2:

\[
\cos\left(2\left(x + \frac{\pi}{3}\right)\right) = -\frac{1}{2}
\]

### Step 2: Solve the cosine equation
The general solution for the cosine equation \(\cos \theta = -\frac{1}{2}\) is:

\[
\theta = \cos^{-1}\left(-\frac{1}{2}\right)
\]

For \(\cos \theta = -\frac{1}{2}\), the solutions in the interval \(0 \leq \theta \leq 2\pi\) are:

\[
\theta = \frac{2\pi}{3} \quad \text{and} \quad \theta = \frac{4\pi}{3}
\]

### Step 3: Substitute back the expression for \(\theta\)
We let:

\[
\theta = 2\left(x + \frac{\pi}{3}\right)
\]

So we have two equations to solve for \(x\):

1. \(2\left(x + \frac{\pi}{3}\right) = \frac{2\pi}{3}\)
2. \(2\left(x + \frac{\pi}{3}\right) = \frac{4\pi}{3}\)

### Step 4: Solve each equation
For the first equation:

\[
2\left(x + \frac{\pi}{3}\right) = \frac{2\pi}{3}
\]

Divide both sides by 2:

\[
x + \frac{\pi}{3} = \frac{\pi}{3}
\]

Subtract \(\frac{\pi}{3}\) from both sides:

\[
x = 0
\]

For the second equation:

\[
2\left(x + \frac{\pi}{3}\right) = \frac{4\pi}{3}
\]

Divide both sides by 2:

\[
x + \frac{\pi}{3} = \frac{2\pi}{3}
\]

Subtract \(\frac{\pi}{3}\) from both sides:

\[
x = \frac{\pi}{3}
\]

### Step 5: General solution
Since the cosine function is periodic, we also need to account for the periodicity of \(2\pi\). The general solutions for \(x\) considering the periodicity are:

\[
x = 0 + n\pi \quad \text{and} \quad x = \frac{\pi}{3} + n\pi \quad \text{where } n \text{ is an integer}
\]

### Step 6: Determine solutions within the interval \(0 \leq x \leq 2\pi\)
Let's identify the values of \(x\) within the interval \(0 \leq x \leq 2\pi\):

For \(x = 0 + n\pi\):
- \(x = 0\)
- \(x = \pi\)
- \(x = 2\pi\) (but \(x = 2\pi\) is not included because the interval is \(0 \leq x < 2\pi\))

For \(x = \frac{\pi}{3} + n\pi\):
- \(x = \frac{\pi}{3}\)
- \(x = \frac{\pi}{3} + \pi = \frac{4\pi}{3}\)

### Final Answer
The solutions in the interval \(0 \leq x < 2\pi\) are:

\[
x = 0, \quad \frac{\pi}{3}, \quad \pi, \quad \text{and} \quad \frac{4\pi}{3}
\]

Would you like any further details or have any questions?

---

Here are some related questions you might find helpful:

1. How can you determine all solutions to trigonometric equations within a specific interval?
2. What are the implications of phase shifts in trigonometric functions on their solutions?
3. How does the period of a trigonometric function affect the number of solutions in a given interval?
4. How can similar techniques be applied to solve sine-based trigonometric equations?
5. What are some common pitfalls when solving trigonometric equations?

**Tip:** Always check the interval carefully when solving trigonometric equations to ensure that all solutions are within the desired range.

Learn how to solve the trigonometric equation 2cos(2(x+3π)) = -1 step-by-step with detailed explanations. This problem involves trigonometric identities and solving cosine equations. Suitable for high school students studying trigonometry.

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