Solve the inequality ` 5+ 2 cos 2x le 3 | 2 sin x - 1|`

Answer

To solve the inequality \( 5 + 2 \cos 2x \leq 3 |2 \sin x - 1| \), we need to consider two cases due to the absolute value.

### Step 1: Break down the absolute value
The absolute value \( |2 \sin x - 1| \) can be split into two cases:
1. \( 2 \sin x - 1 \) when \( 2 \sin x - 1 \geq 0 \) (i.e., \( \sin x \geq \frac{1}{2} \))
2. \( -(2 \sin x - 1) \) when \( 2 \sin x - 1 < 0 \) (i.e., \( \sin x < \frac{1}{2} \))

### Step 2: Case 1 - \( 2 \sin x - 1 \geq 0 \)
In this case, we rewrite the inequality as:
\[ 5 + 2 \cos 2x \leq 3(2 \sin x - 1) \]
This simplifies to:
\[ 5 + 2 \cos 2x \leq 6 \sin x - 3 \]
Rearranging gives:
\[ 2 \cos 2x + 8 \leq 6 \sin x \]
Using the identity \( \cos 2x = 1 - 2 \sin^2 x \), we substitute:
\[ 2(1 - 2 \sin^2 x) + 8 \leq 6 \sin x \]
This simplifies to:
\[ 2 - 4 \sin^2 x + 8 \leq 6 \sin x \]
\[ -4 \sin^2 x - 6 \sin x + 10 \leq 0 \]
Dividing the entire inequality by -2 (and flipping the inequality):
\[ 2 \sin^2 x + 3 \sin x - 5 \geq 0 \]

### Step 3: Factor the quadratic
To factor \( 2 \sin^2 x + 3 \sin x - 5 \), we look for two numbers that multiply to \( -10 \) (product of \( 2 \times -5 \)) and add to \( 3 \). The factors are \( 5 \) and \( -2 \):
\[ (2 \sin x - 1)(\sin x + 5) \geq 0 \]
The critical points are:
1. \( \sin x = \frac{1}{2} \) (from \( 2 \sin x - 1 = 0 \))
2. \( \sin x = -5 \) (not in the range of sine)

### Step 4: Analyze the intervals
The inequality \( (2 \sin x - 1)(\sin x + 5) \geq 0 \) holds when:
- \( \sin x \geq \frac{1}{2} \)

### Step 5: Case 2 - \( 2 \sin x - 1 < 0 \)
Here, we rewrite the inequality as:
\[ 5 + 2 \cos 2x \leq -3(2 \sin x - 1) \]
This simplifies to:
\[ 5 + 2 \cos 2x \leq -6 \sin x + 3 \]
Rearranging gives:
\[ 2 \cos 2x + 8 \leq -6 \sin x \]
Substituting \( \cos 2x = 1 - 2 \sin^2 x \):
\[ 2(1 - 2 \sin^2 x) + 8 \leq -6 \sin x \]
This simplifies to:
\[ 2 - 4 \sin^2 x + 8 \leq -6 \sin x \]
\[ -4 \sin^2 x + 6 \sin x + 10 \leq 0 \]
Dividing by -2 (flipping the inequality):
\[ 2 \sin^2 x - 3 \sin x - 5 \geq 0 \]

### Step 6: Factor the quadratic
Factoring gives:
\[ (2 \sin x + 5)(\sin x - 2) \geq 0 \]
The critical points are:
1. \( \sin x = -\frac{5}{2} \) (not in the range of sine)
2. \( \sin x = 2 \) (not in the range of sine)

### Step 7: Combine results
From Case 1, we found \( \sin x \geq \frac{1}{2} \). The solution set for \( x \) is:
\[ x \geq \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x \leq \frac{5\pi}{6} + 2k\pi \quad \text{for integers } k \]

### Final Answer
The solution to the inequality \( 5 + 2 \cos 2x \leq 3 |2 \sin x - 1| \) is:
\[ x \in \left[ \frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi \right] \quad \text{for integers } k \]

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