in step 2 you made a wrong assumption that 2sin(2x)=-2 and cos(2x)=0. it could be but it's not the only solution. you should factorize the expression 

You're correct that the approach in Step 2 led to an incorrect assumption by focusing on a particular case. Instead, let's solve the equation by factoring the expression properly. Here's the correct approach:

### Step 1: Start with the original equation
\[
2\sin(2x) + \cos(2x) = -2
\]

### Step 2: Move all terms to one side of the equation
\[
2\sin(2x) + \cos(2x) + 2 = 0
\]

### Step 3: Factor the expression
To factor this expression, let's think of it as a combination of terms that can be grouped:

\[
\cos(2x) + 2\sin(2x) + 2 = 0
\]

Notice that we can rewrite this in a more factorable form:
\[
\cos(2x) + 2\sin(2x) = -2
\]

This suggests a strategy where we look for an identity or transformation that can help factor the expression. A suitable approach here is to attempt a trigonometric identity or substitution method to factor. Let's rearrange and try substitution:

### Step 4: Introduce a substitution for \( \sin(2x) \) and \( \cos(2x) \)
Consider substituting \( \sin(2x) = y \) and \( \cos(2x) = \sqrt{1-y^2} \). The equation then becomes:

\[
2y + \sqrt{1-y^2} = -2
\]

Square both sides to eliminate the square root (though we need to be mindful of introducing extraneous solutions):

\[
(2y + 2)^2 = 1 - y^2
\]

Expanding both sides:
\[
4y^2 + 8y + 4 = 1 - y^2
\]

### Step 5: Combine like terms
\[
5y^2 + 8y + 3 = 0
\]

### Step 6: Solve the quadratic equation
Now solve the quadratic equation \( 5y^2 + 8y + 3 = 0 \) using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 5 \), \( b = 8 \), and \( c = 3 \).

\[
y = \frac{-8 \pm \sqrt{64 - 60}}{10} = \frac{-8 \pm \sqrt{4}}{10}
\]
\[
y = \frac{-8 \pm 2}{10}
\]
\[
y = \frac{-6}{10} \quad \text{or} \quad y = \frac{-10}{10}
\]
\[
y = -0.6 \quad \text{or} \quad y = -1
\]

### Step 7: Back-substitute to find \( x \)
- For \( y = -1 \), \( \sin(2x) = -1 \), so:
  \[
  2x = \frac{3\pi}{2} + 2k\pi
  \]
  \[
  x = \frac{3\pi}{4} + k\pi
  \]

- For \( y = -0.6 \), solve:
  \[
  \sin(2x) = -0.6
  \]
  This implies:
  \[
  2x = \arcsin(-0.6) + 2k\pi \quad \text{or} \quad 2x = \pi - \arcsin(-0.6) + 2k\pi
  \]
  \[
  x = \frac{1}{2}\arcsin(-0.6) + k\pi \quad \text{or} \quad x = \frac{\pi}{2} - \frac{1}{2}\arcsin(-0.6) + k\pi
  \]

### Step 8: Combine all solutions
Thus, the solutions to the original equation are:
\[
x = \frac{3\pi}{4} + k\pi
\]
or
\[
x = \frac{1}{2}\arcsin(-0.6) + k\pi
\]
or
\[
x = \frac{\pi}{2} - \frac{1}{2}\arcsin(-0.6) + k\pi
\]
for integer \( k \).

Would you like further clarification or have any questions?

### Further Questions:
1. What is the general solution for \( 2\sin(3x) + \cos(3x) = -2 \)?
2. How does squaring both sides affect the nature of the solutions?
3. Can you find the exact value of \( \arcsin(-0.6) \) in terms of \( \pi \)?
4. What happens if the equation is \( 2\sin(2x) + \cos(2x) = 0 \)?
5. How does changing the coefficients affect the solutions of trigonometric equations?

### Tip:
Always check for extraneous solutions when squaring both sides of an equation.

Learn how to solve the trigonometric equation 2sin(2x) + cos(2x) = -2 using trigonometric identities and substitution methods. This problem explores advanced trigonometry concepts and includes step-by-step solutions and explanations. Suitable for advanced high school students and college-level learners.

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