The general value of `theta` satisfying `sin^(2)theta+sintheta=2` is

Options

A. `npi+(-1)^(n)(pi)/(6)`
B. `2npi+(pi)/(4)`
C. `npi+(-1)^(n)(pi)/(2)`
D. `npi+(-1)^(n)(pi)/(3)`

Explanation

To solve the equation \( \sin^2 \theta + \sin \theta = 2 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ \sin^2 \theta + \sin \theta - 2 = 0 \]
This is a quadratic equation in terms of \( \sin \theta \).

**Hint:** Recognize that this is a standard quadratic form \( ax^2 + bx + c = 0 \).

### Step 2: Factoring the Quadratic
Next, we will factor the quadratic equation:
\[ \sin^2 \theta + \sin \theta - 2 = 0 \]
We can factor this as:
\[ (\sin \theta - 1)(\sin \theta + 2) = 0 \]

**Hint:** Look for two numbers that multiply to \(-2\) and add to \(1\).

### Step 3: Setting Each Factor to Zero
Now, we set each factor equal to zero:
1. \( \sin \theta - 1 = 0 \)
2. \( \sin \theta + 2 = 0 \)

**Hint:** Each factor gives a potential solution for \( \sin \theta \).

### Step 4: Solving Each Equation
1. From \( \sin \theta - 1 = 0 \):
\[ \sin \theta = 1 \]
The general solution for this is:
\[ \theta = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

2. From \( \sin \theta + 2 = 0 \):
\[ \sin \theta = -2 \]
This equation has no solution since the sine function only takes values in the range \([-1, 1]\).

**Hint:** Check the range of the sine function to determine if a solution exists.

### Step 5: Conclusion
The only valid solution comes from \( \sin \theta = 1 \):
\[ \theta = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

### Final Answer
The general value of \( \theta \) satisfying the equation \( \sin^2 \theta + \sin \theta = 2 \) is:
\[ \theta = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

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