If `sinx+cosx=sqrt(y+1/y), y gt 0, x in[0,pi]`, then find the least value of x satisfying the given conditions.

Options

A. `x=pi/4, y=1`
B. `y=0`
C. `y=2`
D. `x=(3pi)/4`

Answer

To solve the problem, we need to find the least value of \( x \) such that:

\[ \sin x + \cos x = \sqrt{y + \frac{1}{y}} \]

where \( y > 0 \) and \( x \in [0, \pi] \).

### Step-by-Step Solution:

1. **Understanding the Range of \( \sin x + \cos x \)**:
The expression \( \sin x + \cos x \) can be rewritten using the identity:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
The maximum value of \( \sin x + \cos x \) occurs when \( x = \frac{\pi}{4} \) and is equal to \( \sqrt{2} \). Therefore:
\[ \sin x + \cos x \leq \sqrt{2} \]

2. **Analyzing the Right-Hand Side**:
The right-hand side \( \sqrt{y + \frac{1}{y}} \) can be minimized by applying the AM-GM inequality:
\[ y + \frac{1}{y} \geq 2 \quad \text{for } y > 0 \]
Thus:
\[ \sqrt{y + \frac{1}{y}} \geq \sqrt{2} \]

3. **Setting the Equality**:
For the equality \( \sin x + \cos x = \sqrt{y + \frac{1}{y}} \) to hold, both sides must equal \( \sqrt{2} \). This occurs when:
\[ y + \frac{1}{y} = 2 \implies y = 1 \]

4. **Finding \( x \)**:
The equality \( \sin x + \cos x = \sqrt{2} \) occurs when:
\[ \sin x = \cos x \implies x = \frac{\pi}{4} \]

5. **Conclusion**:
The least value of \( x \) satisfying the given conditions is:
\[ x = \frac{\pi}{4} \]

### Final Answer:
The least value of \( x \) is \( \frac{\pi}{4} \).