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### 1.1. Giải các phương trình sau

#### a) \(|x^2 + 2| = |x^2 - 1|\)

We have an equation involving absolute values. To solve it, we need to consider different cases based on the expressions inside the absolute value.

1. **Case 1:** \(x^2 + 2 = x^2 - 1\)

   - Simplify: 
     \[
     x^2 + 2 = x^2 - 1 \implies 2 = -1
     \]
     This is a contradiction, so no solution exists in this case.

2. **Case 2:** \(-(x^2 + 2) = x^2 - 1\)

   - Simplify:
     \[
     -x^2 - 2 = x^2 - 1 \implies -2x^2 = 1 \implies x^2 = -\frac{1}{2}
     \]
     Since \(x^2\) cannot be negative, this case has no solution.

3. **Case 3:** \(x^2 + 2 = -(x^2 - 1)\)

   - Simplify:
     \[
     x^2 + 2 = -x^2 + 1 \implies 2x^2 = -1
     \]
     Again, \(x^2\) cannot be negative, so this case also has no solution.

Thus, for part a), there is no solution.

#### b) \(\cos 2x + \cos x + 1 = 0 \quad \text{trên} \quad [0, 2\pi]\)

To solve this trigonometric equation, we use trigonometric identities and simplify the expression:

1. Use the identity \(\cos 2x = 2\cos^2 x - 1\):
   \[
   (2\cos^2 x - 1) + \cos x + 1 = 0
   \]
   Simplify:
   \[
   2\cos^2 x + \cos x = 0
   \]

2. Factor the equation:
   \[
   \cos x (2\cos x + 1) = 0
   \]

3. Solve each factor:
   - \(\cos x = 0\):
     \[
     x = \frac{\pi}{2}, \frac{3\pi}{2}
     \]
   - \(2\cos x + 1 = 0 \implies \cos x = -\frac{1}{2}\):
     \[
     x = \frac{2\pi}{3}, \frac{4\pi}{3}
     \]

Thus, the solutions for part b) are:
\[
x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{2\pi}{3}, \frac{4\pi}{3}
\]

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Do you have any questions about the solutions or need further details?

Here are some related questions you might explore:
1. What are the properties of absolute value equations that allow us to break them into cases?
2. How do we apply trigonometric identities to simplify complex equations?
3. Can the quadratic equation \(2\cos^2 x + \cos x = 0\) be solved using another method, like substitution?
4. What are the general solutions for \(\cos x = 0\) and \(\cos x = -\frac{1}{2}\)?
5. How can contradictions arise in algebraic equations involving squares and absolute values?

**Tip:** Always check for contradictions when solving absolute value equations, as some cases

a) |x^2 + 2| = |x^2 - 1|
b) cos 2x + cos x + 1 = 0 trên [0, 2π]

Learn how to solve the absolute value equation |x^2 + 2| = |x^2 - 1| and the trigonometric equation cos 2x + cos x + 1 = 0 on the interval [0, 2π]. This tutorial covers key algebraic and trigonometric concepts, such as the absolute value property and trigonometric identities. Suitable for students in Grades 10-12.

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