The number of values of x which satify the equation `cos^(2)x =1` and `x^(2) le 20` are _________

Answer

To solve the equation \( \cos^2 x = 1 \) under the condition \( x^2 \leq 20 \), we will follow these steps:

### Step 1: Solve the equation \( \cos^2 x = 1 \)

The equation \( \cos^2 x = 1 \) implies that \( \cos x = \pm 1 \).

### Step 2: Find the general solutions for \( \cos x = 1 \) and \( \cos x = -1 \)

1. For \( \cos x = 1 \):
- The solutions are given by:
\[ x = 2n\pi \quad (n \in \mathbb{Z}) \]

2. For \( \cos x = -1 \):
- The solutions are given by:
\[ x = (2n + 1)\pi \quad (n \in \mathbb{Z}) \]

### Step 3: Combine the solutions

The complete set of solutions can be expressed as:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]

### Step 4: Apply the condition \( x^2 \leq 20 \)

Now we need to find the integer values of \( n \) such that:
\[ (n\pi)^2 \leq 20 \]
This simplifies to:
\[ n^2 \leq \frac{20}{\pi^2} \]

### Step 5: Calculate \( \frac{20}{\pi^2} \)

Using \( \pi \approx 3.14 \):
\[ \pi^2 \approx 9.86 \quad \Rightarrow \quad \frac{20}{\pi^2} \approx \frac{20}{9.86} \approx 2.02 \]

### Step 6: Determine the integer values of \( n \)

The inequality \( n^2 \leq 2.02 \) gives us:
\[ n \in \{-1, 0, 1\} \]
This means \( n \) can take the values -1, 0, and 1.

### Step 7: Find corresponding values of \( x \)

1. For \( n = -1 \):
\[ x = -\pi \approx -3.14 \]
2. For \( n = 0 \):
\[ x = 0 \]
3. For \( n = 1 \):
\[ x = \pi \approx 3.14 \]

### Step 8: Count the valid \( x \) values

The values of \( x \) that satisfy both conditions are:
- \( -\pi \)
- \( 0 \)
- \( \pi \)

Thus, there are a total of **3 values** of \( x \) that satisfy the equation \( \cos^2 x = 1 \) and \( x^2 \leq 20 \).

### Final Answer:
The number of values of \( x \) which satisfy the equation is **3**.

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