`2 sin^2 a - 3 sin a - 2 = 0` then `a=`

Options

A. `n pi + (-1)^n pi/2`
B. `n pi +(-1)^n pi//6`
C. `n pi + (-1)^n (7pi)/6`
D. `n pi - (-1)^n pi/6`

Answer

To solve the equation \(2 \sin^2 a - 3 \sin a - 2 = 0\), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ 2 \sin^2 a - 3 \sin a - 2 = 0 \]

### Step 2: Factoring the Quadratic
Next, we can factor the quadratic equation. We need to find two numbers that multiply to \(2 \times -2 = -4\) and add up to \(-3\). The numbers \(-4\) and \(1\) work:
\[ 2 \sin^2 a - 4 \sin a + \sin a - 2 = 0 \]
Now, we can group the terms:
\[ (2 \sin^2 a - 4 \sin a) + (\sin a - 2) = 0 \]

### Step 3: Taking Common Factors
Now, we factor by grouping:
\[ 2 \sin a (\sin a - 2) + 1 (\sin a - 2) = 0 \]
This gives us:
\[ (2 \sin a + 1)(\sin a - 2) = 0 \]

### Step 4: Setting Each Factor to Zero
Now we set each factor to zero:
1. \(2 \sin a + 1 = 0\)
2. \(\sin a - 2 = 0\)

### Step 5: Solving the First Factor
From the first factor:
\[ 2 \sin a + 1 = 0 \implies \sin a = -\frac{1}{2} \]

### Step 6: Solving the Second Factor
From the second factor:
\[ \sin a - 2 = 0 \implies \sin a = 2 \]
However, since the sine function only takes values in the range \([-1, 1]\), this equation has no valid solutions.

### Step 7: Finding Angles for \(\sin a = -\frac{1}{2}\)
Now we need to find the angles \(a\) where \(\sin a = -\frac{1}{2}\). The sine function is negative in the third and fourth quadrants:
- In the third quadrant:
\[ a = \frac{7\pi}{6} \]
- In the fourth quadrant:
\[ a = \frac{11\pi}{6} \]

### Step 8: General Solution
The general solutions for \(a\) can be expressed as:
\[ a = n\pi + \frac{7\pi}{6} \quad \text{and} \quad a = n\pi + \frac{11\pi}{6}, \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the solutions for \(a\) are:
\[ a = n\pi + \frac{7\pi}{6} \quad \text{and} \quad a = n\pi + \frac{11\pi}{6}, \quad n \in \mathbb{Z} \]