Given a continuous function $ f(x)$ such that $ \int_0^{2\pi} f(x)\ dx \equal{} 0$.

Let $ S(x) \equal{} A_0 \plus{} A_1\cos x \plus{} B_1\sin x$, find constant numbers $ A_0,\ A_1$ and $ B_1$ for which $ \int_0^{2\pi} \{f(x) \minus{} S(x)\}^2\ dx$ is minimized.
To find the constants \( A_0, A_1, \) and \( B_1 \) that minimize the integral \( \int_0^{2\pi} \{f(x) - S(x)\}^2 \, dx \), where \( S(x) = A_0 + A_1 \cos x + B_1 \sin x \), we need to minimize the function \( F(a, b, c) = \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x]^2 \, dx \).

1. **Set up the function to minimize:**
   \[
   F(a, b, c) = \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x]^2 \, dx
   \]

2. **Compute the partial derivatives and set them to zero:**
   \[
   \frac{\partial F}{\partial a} = 0, \quad \frac{\partial F}{\partial b} = 0, \quad \frac{\partial F}{\partial c} = 0
   \]

3. **Calculate \(\frac{\partial F}{\partial a}\):**
   \[
   \frac{\partial F}{\partial a} = \frac{\partial}{\partial a} \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x]^2 \, dx
   \]
   Using the chain rule:
   \[
   \int_0^{2\pi} 2 [f(x) - a - b \cos x - c \sin x] (-1) \, dx = 0
   \]
   Simplifying:
   \[
   -2 \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x] \, dx = 0
   \]
   Since \(\int_0^{2\pi} f(x) \, dx = 0\):
   \[
   -2 \left( \int_0^{2\pi} f(x) \, dx - a \int_0^{2\pi} 1 \, dx - b \int_0^{2\pi} \cos x \, dx - c \int_0^{2\pi} \sin x \, dx \right) = 0
   \]
   \[
   -2 \left( 0 - a \cdot 2\pi - b \cdot 0 - c \cdot 0 \right) = 0
   \]
   \[
   a = 0
   \]

4. **Calculate \(\frac{\partial F}{\partial b}\):**
   \[
   \frac{\partial F}{\partial b} = \frac{\partial}{\partial b} \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x]^2 \, dx
   \]
   Using the chain rule:
   \[
   \int_0^{2\pi} 2 [f(x) - a - b \cos x - c \sin x] (-\cos x) \, dx = 0
   \]
   Simplifying:
   \[
   -2 \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x] \cos x \, dx = 0
   \]
   Since \(a = 0\):
   \[
   -2 \left( \int_0^{2\pi} f(x) \cos x \, dx - b \int_0^{2\pi} \cos^2 x \, dx - c \int_0^{2\pi} \cos x \sin x \, dx \right) = 0
   \]
   \[
   -2 \left( \int_0^{2\pi} f(x) \cos x \, dx - b \cdot \pi - c \cdot 0 \right) = 0
   \]
   \[
   b = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos x \, dx
   \]

5. **Calculate \(\frac{\partial F}{\partial c}\):**
   \[
   \frac{\partial F}{\partial c} = \frac{\partial}{\partial c} \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x]^2 \, dx
   \]
   Using the chain rule:
   \[
   \int_0^{2\pi} 2 [f(x) - a - b \cos x - c \sin x] (-\sin x) \, dx = 0
   \]
   Simplifying:
   \[
   -2 \int_0^{2\pi} [f(x) - a - b \cos x - c \sin x] \sin x \, dx = 0
   \]
   Since \(a = 0\):
   \[
   -2 \left( \int_0^{2\pi} f(x) \sin x \, dx - b \int_0^{2\pi} \cos x \sin x \, dx - c \int_0^{2\pi} \sin^2 x \, dx \right) = 0
   \]
   \[
   -2 \left( \int_0^{2\pi} f(x) \sin x \, dx - b \cdot 0 - c \cdot \pi \right) = 0
   \]
   \[
   c = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin x \, dx
   \]

Thus, the constants \( A_0, A_1, \) and \( B_1 \) that minimize the integral are:
\[
A_0 = 0, \quad A_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos x \, dx, \quad B_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin x \, dx
\]

The final answer is \( \boxed{ A_0 = 0, A_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos x \, dx, B_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin x \, dx } \)