If the solutions for `theta` from the equation `sin^(2) theta-2sin theta + lambda = 0 ` lie in ` cup_( n in z)(2 n pi - pi/6, (2n+1) pi + (pi)/(6))`. Then, find the possible set values of `lambda` .

Answer

To solve the problem, we need to analyze the given equation and the conditions provided for the solutions of \(\theta\).

### Step 1: Rewrite the given equation
The equation given is:
\[ \sin^2 \theta - 2 \sin \theta + \lambda = 0 \]
This is a quadratic equation in terms of \(\sin \theta\).

### Step 2: Identify the quadratic formula
For a quadratic equation of the form \(ax^2 + bx + c = 0\), the solutions can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \(a = 1\), \(b = -2\), and \(c = \lambda\).

### Step 3: Find the discriminant
The discriminant (\(D\)) of the quadratic equation must be non-negative for real solutions to exist:
\[ D = b^2 - 4ac = (-2)^2 - 4(1)(\lambda) = 4 - 4\lambda \]
For real solutions, we need:
\[ 4 - 4\lambda \geq 0 \]
This simplifies to:
\[ 1 - \lambda \geq 0 \quad \Rightarrow \quad \lambda \leq 1 \]

### Step 4: Analyze the range of \(\sin \theta\)
The solutions for \(\theta\) lie in the intervals:
\[ \bigcup_{n \in \mathbb{Z}} \left(2n\pi - \frac{\pi}{6}, (2n+1)\pi + \frac{\pi}{6}\right) \]
This implies that \(\sin \theta\) must take values in the range:
\[ -\frac{1}{2} \leq \sin \theta \leq 1 \]

### Step 5: Substitute \(\sin \theta\) into the quadratic equation
Let \(x = \sin \theta\). The equation becomes:
\[ x^2 - 2x + \lambda = 0 \]
The roots of this equation must lie within the range of \(\sin \theta\), which is \([-1/2, 1]\).

### Step 6: Find the maximum and minimum values of \(\lambda\)
The vertex of the parabola described by the quadratic equation \(x^2 - 2x + \lambda\) occurs at:
\[ x = \frac{-b}{2a} = \frac{2}{2} = 1 \]
Substituting \(x = 1\) into the equation gives:
\[ 1^2 - 2(1) + \lambda = 0 \quad \Rightarrow \quad \lambda = 1 \]

Now, substituting \(x = -\frac{1}{2}\):
\[ \left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) + \lambda = 0 \]
This gives:
\[ \frac{1}{4} + 1 + \lambda = 0 \quad \Rightarrow \quad \lambda = -\frac{5}{4} \]

### Step 7: Combine results to find the range of \(\lambda\)
From the analysis, we have:
\[ -\frac{5}{4} < \lambda \leq 1 \]
Thus, the possible set of values for \(\lambda\) is:
\[ \lambda \in \left(-\frac{5}{4}, 1\right] \]

### Final Answer
The possible set values of \(\lambda\) is:
\[ \lambda \in \left(-\frac{5}{4}, 1\right] \]