The inequality ` 4 sin 3x + 5 ge 4 cos 2x + 5 sin x ` true for ` x in `

Options

A. `[- pi , (3pi)/(2)]`
B. `[ - (pi)/(2) , (pi)/(2)]`
C. `[(5pi)/(8), (13 pi)/(8)] `
D. `[(23 pi)/(14) , (41 pi)/(14)]`

Answer

To solve the inequality \( 4 \sin 3x + 5 \geq 4 \cos 2x + 5 \sin x \), we will follow these steps:

### Step 1: Rewrite the inequality
Start with the given inequality:
\[ 4 \sin 3x + 5 \geq 4 \cos 2x + 5 \sin x \]

### Step 2: Use trigonometric identities
We know the following trigonometric identities:
- \( \sin 3x = 3 \sin x - 4 \sin^3 x \)
- \( \cos 2x = 1 - 2 \sin^2 x \)

Substituting these identities into the inequality gives:
\[ 4(3 \sin x - 4 \sin^3 x) + 5 \geq 4(1 - 2 \sin^2 x) + 5 \sin x \]

### Step 3: Simplify the inequality
Expanding both sides:
\[ 12 \sin x - 16 \sin^3 x + 5 \geq 4 - 8 \sin^2 x + 5 \sin x \]
Combine like terms:
\[ 12 \sin x - 16 \sin^3 x + 5 - 4 + 8 \sin^2 x - 5 \sin x \geq 0 \]
This simplifies to:
\[ -16 \sin^3 x + 8 \sin^2 x + 7 \sin x + 1 \geq 0 \]

### Step 4: Rearrange the inequality
Rearranging gives:
\[ 16 \sin^3 x - 8 \sin^2 x - 7 \sin x - 1 \leq 0 \]

### Step 5: Factor the polynomial
We can factor out \( \sin x - 1 \):
\[ (16 \sin^2 x + 8 \sin x + 1)(\sin x - 1) \leq 0 \]

### Step 6: Analyze the factors
The quadratic \( 16 \sin^2 x + 8 \sin x + 1 \) is always positive since its discriminant \( 8^2 - 4 \times 16 \times 1 = 64 - 64 = 0 \) indicates it has a double root and does not change sign. The only factor we need to consider is:
\[ \sin x - 1 \leq 0 \]

### Step 7: Solve the inequality
This simplifies to:
\[ \sin x \leq 1 \]
Since the sine function achieves its maximum value of 1, this inequality holds for all \( x \).

### Step 8: Determine the range of \( x \)
Since \( \sin x \) can take values from -1 to 1, the inequality is satisfied for all \( x \) in the interval:
\[ x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]

### Final Answer
Thus, the inequality \( 4 \sin 3x + 5 \geq 4 \cos 2x + 5 \sin x \) is true for:
\[ x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]

---