Determine the critical points of the autonomous differential equation $$ \frac{d y}{d x}=\left(\frac{1}{\pi} y+1\right)-\cos \frac{y}{2} $$ $$ -\frac{\pi}{2}, 0, \frac{\pi}{2} $$ $$ -\pi, 0, \pi $$ $$ 0,-\pi $$ $$ 0, \pi $$ $$ -\frac{\pi}{2}, \frac{\pi}{2} $$

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To determine the critical points of the autonomous differential equation

$$
\frac{d y}{d x} = \left(\frac{1}{\pi} y + 1\right) - \cos\left(\frac{y}{2}\right),
$$

we need to find the values of $$ y $$ where $$ \frac{d y}{d x} = 0 $$. Setting the derivative equal to zero gives:

$$
\frac{1}{\pi} y + 1 - \cos\left(\frac{y}{2}\right) = 0.
$$

This can be rearranged to:

$$
\frac{y}{\pi} + 1 = \cos\left(\frac{y}{2}\right).
$$

### Step 1: Determine the Range of Possible Solutions

The cosine function has a range of $$[-1, 1]$$. Therefore, for real solutions:

$$
-1 \leq \frac{y}{\pi} + 1 \leq 1.
$$

Solving the inequalities:

$$
-2 \leq \frac{y}{\pi} \leq 0 \quad \Rightarrow \quad -2\pi \leq y \leq 0.
$$

Thus, we only need to consider $$ y $$ values within $$[-2\pi, 0]$$.

### Step 2: Identify Critical Points

Let's test potential critical points within the specified range:

1. **$$ y = 0 $$:**
   $$
   \frac{0}{\pi} + 1 = 1, \quad \cos\left(0\right) = 1.
   $$
   Hence, $$ y = 0 $$ is a critical point.

2. **$$ y = -\pi $$:**
   $$
   \frac{-\pi}{\pi} + 1 = 0, \quad \cos\left(-\frac{\pi}{2}\right) = 0.
   $$
   Thus, $$ y = -\pi $$ is also a critical point.

3. **$$ y = -2\pi $$:**
   $$
   \frac{-2\pi}{\pi} + 1 = -1, \quad \cos\left(-\pi\right) = -1.
   $$
   So, $$ y = -2\pi $$ is another critical point.

However, the options provided do not include $$ y = -2\pi $$. Among the given choices, **$$ y = 0 $$** and **$$ y = -\pi $$** are the correct critical points.

### Conclusion

The critical points of the differential equation are $$ y = 0 $$ and $$ y = -\pi $$.

**Answer:** $$0, -\pi$$
The critical points are $$ y = 0 $$ and $$ y = -\pi $$.

Determine the critical points of the autonomous differential equation

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