For what value of `k` the equation `sinx+cos(k+x)+cos(k-x)=2` has real solutions?

Explanation

To solve the equation \( \sin x + \cos(k + x) + \cos(k - x) = 2 \) for the value of \( k \) that allows for real solutions, we can follow these steps:

### Step 1: Simplify the Equation
We start with the equation:
\[ \sin x + \cos(k + x) + \cos(k - x) = 2 \]
Using the cosine addition formula, we can rewrite \( \cos(k + x) + \cos(k - x) \):
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