In `[-pi, pi]`, solution of inequality………. `cos 2x+5cos x +3 ge 0` is

Options

A. `[-pi, pi]`
B. `[(-5pi)/(6), (pi)/(6)]`
C. `[0, pi]`
D. `[(-2pi)/(3), (2pi)/(3)]`

Explanation

To solve the inequality \( \cos 2x + 5 \cos x + 3 \geq 0 \) in the interval \([-π, π]\), we can follow these steps:

### Step 1: Rewrite \( \cos 2x \)
Using the double angle formula for cosine, we have:
\[ \cos 2x = 2 \cos^2 x - 1 \]
Substituting this into the inequality gives:
\[ 2 \cos^2 x - 1 + 5 \cos x + 3 \geq 0 \]

### Step 2: Simplify the inequality
Combine like terms:
\[ 2 \cos^2 x + 5 \cos x + 2 \geq 0 \]

### Step 3: Let \( y = \cos x \)
Now, we can rewrite the inequality in terms of \( y \):
\[ 2y^2 + 5y + 2 \geq 0 \]

### Step 4: Factor the quadratic
We need to factor the quadratic expression \( 2y^2 + 5y + 2 \). To do this, we can find the roots using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = 5, c = 2 \).

Calculating the discriminant:
\[ b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \]
Now, substituting into the quadratic formula:
\[ y = \frac{-5 \pm \sqrt{9}}{2 \cdot 2} = \frac{-5 \pm 3}{4} \]
This gives us two roots:
\[ y_1 = \frac{-2}{4} = -\frac{1}{2}, \quad y_2 = \frac{-8}{4} = -2 \]

### Step 5: Analyze the quadratic
The quadratic \( 2y^2 + 5y + 2 \) opens upwards (since the coefficient of \( y^2 \) is positive). The roots are \( y = -\frac{1}{2} \) and \( y = -2 \). Since \( y = \cos x \) must be in the range \([-1, 1]\), we only consider \( y = -\frac{1}{2} \).

### Step 6: Determine the intervals
The quadratic is non-negative when:
1. \( y \leq -2 \) (not applicable since \( y \) cannot be less than -1)
2. \( y \geq -\frac{1}{2} \)

Thus, we need to solve:
\[ \cos x \geq -\frac{1}{2} \]

### Step 7: Find the values of \( x \)
The cosine function is greater than or equal to \(-\frac{1}{2}\) in the intervals:
\[ x \in [-\frac{2\pi}{3}, \frac{2\pi}{3}] \quad \text{(within the interval } [-\pi, \pi]\text{)} \]

### Final Solution
Thus, the solution of the inequality \( \cos 2x + 5 \cos x + 3 \geq 0 \) in the interval \([-π, π]\) is:
\[ x \in \left[-\frac{2\pi}{3}, \frac{2\pi}{3}\right] \]