If `(sin alpha)^(x)+(cos alpha)^(x) ge 1,0 lt a lt (pi)/(2)` then

Options

A. `n in [2,oo)`
B. `(-oo,2]`
C. `n in [-1,1]`
D. None of these

Answer

To solve the inequality \((\sin \alpha)^x + (\cos \alpha)^x \geq 1\) for \(0 < \alpha < \frac{\pi}{2}\), we will follow these steps:

### Step 1: Understand the inequality
We start with the inequality:
\[ (\sin \alpha)^x + (\cos \alpha)^x \geq 1 \]
for \(0 < \alpha < \frac{\pi}{2}\).

### Step 2: Use the identity
We know from trigonometric identities that:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]
We can use this identity to compare our expression with 1.

### Step 3: Rearranging the inequality
Rearranging the inequality gives us:
\[ (\sin \alpha)^x + (\cos \alpha)^x - 1 \geq 0 \]
Let’s define a function \(f(x)\):
\[ f(x) = (\sin \alpha)^x + (\cos \alpha)^x - 1 \]
We need to find the values of \(x\) for which \(f(x) \geq 0\).

### Step 4: Analyze the behavior of \(f(x)\)
Since both \(\sin \alpha\) and \(\cos \alpha\) are positive in the interval \(0 < \alpha < \frac{\pi}{2}\), we can analyze the function \(f(x)\).

### Step 5: Consider the case when \(x = 2\)
Let’s evaluate \(f(2)\):
\[ f(2) = (\sin \alpha)^2 + (\cos \alpha)^2 - 1 = 0 \]
This means that at \(x = 2\), the function equals 0.

### Step 6: Consider \(x < 2\)
Now, if \(x < 2\), we need to check if \(f(x) \geq 0\). As \(x\) decreases from 2, both \((\sin \alpha)^x\) and \((\cos \alpha)^x\) increase because the base values are less than 1. Thus, \(f(x)\) will be positive.

### Step 7: Consider \(x > 2\)
If \(x > 2\), \((\sin \alpha)^x\) and \((\cos \alpha)^x\) will decrease, leading \(f(x)\) to become negative. Hence, \(f(x) < 0\) for \(x > 2\).

### Conclusion
From the analysis, we conclude that:
\[ x < 2 \]
Thus, the solution set is:
\[ x \in (-\infty, 2) \]

### Final Answer
The correct option is:
\[ (-\infty, 2) \]