The solution set of `x in(-pi,pi)` for the inequality `sin 2x+1 le cosx+2sin x` is

Options

A. `x in [0,(pi)/(6)]`
B. `x in [(pi)/(6),(5pi)/(6)]uu{0}`
C. `x in [-(pi)/(6),(5pi)/(6)]`
D. `x in [-(pi)/(2),(pi)/(2)]`

Answer

To solve the inequality \( \sin 2x + 1 \leq \cos x + 2 \sin x \) for \( x \in (-\pi, \pi) \), we can follow these steps:

### Step 1: Rewrite the Inequality
Start with the given inequality:
\[ \sin 2x + 1 \leq \cos x + 2 \sin x \]
Using the double angle identity for sine, \( \sin 2x = 2 \sin x \cos x \), we can rewrite the inequality:
\[ 2 \sin x \cos x + 1 \leq \cos x + 2 \sin x \]

### Step 2: Rearrange the Inequality
Now, move all terms to one side:
\[ 2 \sin x \cos x - \cos x - 2 \sin x + 1 \leq 0 \]
This simplifies to:
\[ \cos x (2 \sin x - 1) - 2 \sin x + 1 \leq 0 \]

### Step 3: Factor the Expression
We can factor the left-hand side:
\[ (2 \sin x - 1)(\cos x - 1) \leq 0 \]

### Step 4: Find Critical Points
To find the critical points, we set each factor to zero:
1. \( 2 \sin x - 1 = 0 \) gives \( \sin x = \frac{1}{2} \) which leads to:
- \( x = \frac{\pi}{6} \)
- \( x = \frac{5\pi}{6} \)

2. \( \cos x - 1 = 0 \) gives \( \cos x = 1 \) which leads to:
- \( x = 0 \)

### Step 5: Determine Intervals
The critical points divide the interval \( (-\pi, \pi) \) into segments:
- \( (-\pi, \frac{\pi}{6}) \)
- \( (\frac{\pi}{6}, 0) \)
- \( (0, \frac{5\pi}{6}) \)
- \( (\frac{5\pi}{6}, \pi) \)

### Step 6: Test Each Interval
We need to test the sign of \( (2 \sin x - 1)(\cos x - 1) \) in each interval:
1. **Interval** \( (-\pi, \frac{\pi}{6}) \): Choose \( x = -\frac{\pi}{3} \)
- \( 2 \sin(-\frac{\pi}{3}) - 1 < 0 \) and \( \cos(-\frac{\pi}{3}) - 1 < 0 \) → Positive
2. **Interval** \( (\frac{\pi}{6}, 0) \): Choose \( x = -\frac{\pi}{12} \)
- \( 2 \sin(-\frac{\pi}{12}) - 1 < 0 \) and \( \cos(-\frac{\pi}{12}) - 1 < 0 \) → Positive
3. **Interval** \( (0, \frac{5\pi}{6}) \): Choose \( x = \frac{\pi}{3} \)
- \( 2 \sin(\frac{\pi}{3}) - 1 > 0 \) and \( \cos(\frac{\pi}{3}) - 1 < 0 \) → Negative
4. **Interval** \( (\frac{5\pi}{6}, \pi) \): Choose \( x = \frac{3\pi}{4} \)
- \( 2 \sin(\frac{3\pi}{4}) - 1 > 0 \) and \( \cos(\frac{3\pi}{4}) - 1 < 0 \) → Negative

### Step 7: Combine Results
The inequality \( (2 \sin x - 1)(\cos x - 1) \leq 0 \) is satisfied in the intervals where the product is negative or zero:
- The critical points \( x = 0, \frac{\pi}{6}, \frac{5\pi}{6} \) are included.

Thus, the solution set is:
\[ x \in [0, \frac{5\pi}{6}] \]