Let $x,\ y$ be real numbers. For a function $f(t)=x\sin t+y\cos t$, draw the domain of the points $(x,\ y)$ for which the following inequality holds.
\[\left|\int_{-\pi}^{\pi} f(t)\cos t\ dt\right|\leq \int_{-\pi}^{\pi} \{f(t)\}^2dt.\]
1. First, we need to evaluate the integral \(\int_{-\pi}^\pi f(t) \cos t \, dt\). Given \(f(t) = x \sin t + y \cos t\), we have:
   \[
   \int_{-\pi}^\pi f(t) \cos t \, dt = \int_{-\pi}^\pi (x \sin t + y \cos t) \cos t \, dt
   \]
   This can be split into two integrals:
   \[
   \int_{-\pi}^\pi x \sin t \cos t \, dt + \int_{-\pi}^\pi y \cos^2 t \, dt
   \]

2. Evaluate the first integral:
   \[
   \int_{-\pi}^\pi x \sin t \cos t \, dt = x \int_{-\pi}^\pi \sin t \cos t \, dt
   \]
   Using the trigonometric identity \(\sin t \cos t = \frac{1}{2} \sin 2t\), we get:
   \[
   x \int_{-\pi}^\pi \frac{1}{2} \sin 2t \, dt = \frac{x}{2} \int_{-\pi}^\pi \sin 2t \, dt
   \]
   Since \(\sin 2t\) is an odd function, its integral over \([- \pi, \pi]\) is zero:
   \[
   \frac{x}{2} \int_{-\pi}^\pi \sin 2t \, dt = 0
   \]

3. Evaluate the second integral:
   \[
   \int_{-\pi}^\pi y \cos^2 t \, dt = y \int_{-\pi}^\pi \cos^2 t \, dt
   \]
   Using the identity \(\cos^2 t = \frac{1 + \cos 2t}{2}\), we get:
   \[
   y \int_{-\pi}^\pi \frac{1 + \cos 2t}{2} \, dt = \frac{y}{2} \int_{-\pi}^\pi (1 + \cos 2t) \, dt
   \]
   The integral of \(\cos 2t\) over \([- \pi, \pi]\) is zero, so:
   \[
   \frac{y}{2} \int_{-\pi}^\pi 1 \, dt = \frac{y}{2} \cdot 2\pi = y\pi
   \]

4. Therefore, we have:
   \[
   \int_{-\pi}^\pi f(t) \cos t \, dt = y\pi
   \]

5. Next, we need to evaluate the integral \(\int_{-\pi}^\pi [f(t)]^2 \, dt\):
   \[
   \int_{-\pi}^\pi [f(t)]^2 \, dt = \int_{-\pi}^\pi (x \sin t + y \cos t)^2 \, dt
   \]
   Expanding the square, we get:
   \[
   \int_{-\pi}^\pi (x^2 \sin^2 t + 2xy \sin t \cos t + y^2 \cos^2 t) \, dt
   \]

6. Evaluate each term separately:
   \[
   \int_{-\pi}^\pi x^2 \sin^2 t \, dt = x^2 \int_{-\pi}^\pi \sin^2 t \, dt
   \]
   Using the identity \(\sin^2 t = \frac{1 - \cos 2t}{2}\), we get:
   \[
   x^2 \int_{-\pi}^\pi \frac{1 - \cos 2t}{2} \, dt = \frac{x^2}{2} \int_{-\pi}^\pi (1 - \cos 2t) \, dt = \frac{x^2}{2} \cdot 2\pi = x^2 \pi
   \]

7. The second term:
   \[
   \int_{-\pi}^\pi 2xy \sin t \cos t \, dt = 2xy \int_{-\pi}^\pi \sin t \cos t \, dt = 0
   \]
   (as shown earlier)

8. The third term:
   \[
   \int_{-\pi}^\pi y^2 \cos^2 t \, dt = y^2 \int_{-\pi}^\pi \cos^2 t \, dt
   \]
   Using the identity \(\cos^2 t = \frac{1 + \cos 2t}{2}\), we get:
   \[
   y^2 \int_{-\pi}^\pi \frac{1 + \cos 2t}{2} \, dt = \frac{y^2}{2} \int_{-\pi}^\pi (1 + \cos 2t) \, dt = \frac{y^2}{2} \cdot 2\pi = y^2 \pi
   \]

9. Therefore, we have:
   \[
   \int_{-\pi}^\pi [f(t)]^2 \, dt = x^2 \pi + y^2 \pi = \pi (x^2 + y^2)
   \]

10. The given inequality is:
    \[
    \left| \int_{-\pi}^\pi f(t) \cos t \, dt \right| \leq \int_{-\pi}^\pi [f(t)]^2 \, dt
    \]
    Substituting the evaluated integrals, we get:
    \[
    |y\pi| \leq \pi (x^2 + y^2)
    \]
    Dividing both sides by \(\pi\), we obtain:
    \[
    |y| \leq x^2 + y^2
    \]

11. To find the domain of \((x, y)\) that satisfies this inequality, we rearrange it:
    \[
    x^2 + y^2 \geq |y|
    \]
    This can be written as:
    \[
    x^2 + y^2 \geq y \quad \text{and} \quad x^2 + y^2 \geq -y
    \]

12. For \(x^2 + y^2 \geq y\):
    \[
    x^2 + y^2 - y \geq 0 \implies x^2 + (y - \frac{1}{2})^2 \geq \frac{1}{4}
    \]

13. For \(x^2 + y^2 \geq -y\):
    \[
    x^2 + y^2 + y \geq 0 \implies x^2 + (y + \frac{1}{2})^2 \geq \frac{1}{4}
    \]

14. The domain of \((x, y)\) is the union of the regions defined by these inequalities. This represents the area outside the circles centered at \((0, \pm \frac{1}{2})\) with radius \(\frac{1}{2}\).

The final answer is \( \boxed{ x^2 + \left(y - \frac{1}{2}\right)^2 \geq \frac{1}{4} } \) and \(x^2 + \left(y + \frac{1}{2}\right)^2 \geq \frac{1}{4}\).