if `2(2 sin theta -1)x^(2)+8x+4(1+sin theta)geforall x in R` then find the general solution of `theta`

Answer

To solve the inequality \( 2(2 \sin \theta - 1)x^2 + 8x + 4(1 + \sin \theta) \geq 0 \) for all \( x \in \mathbb{R} \), we will analyze the quadratic expression in terms of \( x \).

### Step 1: Identify the quadratic equation
The given expression can be rewritten as:
\[ f(x) = 2(2 \sin \theta - 1)x^2 + 8x + 4(1 + \sin \theta) \]
This is a quadratic equation in the form \( ax^2 + bx + c \) where:
- \( a = 2(2 \sin \theta - 1) \)
- \( b = 8 \)
- \( c = 4(1 + \sin \theta) \)

### Step 2: Condition for the quadratic to be non-negative
For the quadratic \( f(x) \) to be non-negative for all \( x \in \mathbb{R} \), two conditions must be satisfied:
1. The leading coefficient \( a \) must be non-negative:
\[ 2(2 \sin \theta - 1) \geq 0 \]
This simplifies to:
\[ 2 \sin \theta - 1 \geq 0 \implies \sin \theta \geq \frac{1}{2} \]

2. The discriminant \( D \) must be less than or equal to zero:
\[ D = b^2 - 4ac \leq 0 \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = 8^2 - 4 \cdot 2(2 \sin \theta - 1) \cdot 4(1 + \sin \theta) \leq 0 \]
Simplifying this gives:
\[ 64 - 32(2 \sin \theta - 1)(1 + \sin \theta) \leq 0 \]

### Step 3: Simplify the discriminant condition
Expanding the discriminant condition:
\[ 64 - 32(2 \sin \theta + 2 \sin^2 \theta - 1 - \sin \theta) \leq 0 \]
This simplifies to:
\[ 64 - 32(2 \sin^2 \theta + \sin \theta + 1) \leq 0 \]
\[ 64 - 64 \sin^2 \theta - 32 \sin \theta - 32 \leq 0 \]
\[ 32 - 64 \sin^2 \theta - 32 \sin \theta \leq 0 \]
\[ -64 \sin^2 \theta - 32 \sin \theta + 32 \leq 0 \]
Dividing through by -32 (reversing the inequality):
\[ 2 \sin^2 \theta + \sin \theta - 1 \geq 0 \]

### Step 4: Factor the quadratic
Factoring the quadratic:
\[ (2 \sin \theta + 2)(\sin \theta - 1) \geq 0 \]
The roots of the equation are:
\[ \sin \theta = -1 \quad \text{and} \quad \sin \theta = \frac{1}{2} \]

### Step 5: Analyze the intervals
To find the intervals where the product is non-negative, we analyze the sign of the factors:
1. \( \sin \theta \leq -1 \) (not possible)
2. \( -1 < \sin \theta < \frac{1}{2} \) (negative)
3. \( \sin \theta = \frac{1}{2} \) (zero)
4. \( \sin \theta > \frac{1}{2} \) (positive)

Thus, the valid intervals for \( \sin \theta \) are:
\[ \sin \theta \geq \frac{1}{2} \]

### Step 6: General solution for \( \theta \)
The general solution for \( \theta \) where \( \sin \theta \geq \frac{1}{2} \) is:
\[ \theta = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \]