To solve the equation \( \cos^2 x + \cos^2 2x = 2 \) in the interval \([0, 20]\), we will follow these steps:
### Step 1: Analyze the equation
The equation \( \cos^2 x + \cos^2 2x = 2 \) implies that both \( \cos^2 x \) and \( \cos^2 2x \) must equal 1, since the maximum value of \( \cos^2 \) is 1.
### Step 2: Set up the individual equations
From \( \cos^2 x = 1 \), we have:
\[
\cos x = \pm 1
\]
This occurs when:
\[
x = n\pi \quad (n \in \mathbb{Z})
\]
From \( \cos^2 2x = 1 \), we have:
\[
\cos 2x = \pm 1
\]
This occurs when:
\[
2x = m\pi \quad (m \in \mathbb{Z}) \implies x = \frac{m\pi}{2}
\]
### Step 3: Find the common solutions
We need to find values of \( x \) that satisfy both conditions:
1. \( x = n\pi \)
2. \( x = \frac{m\pi}{2} \)
### Step 4: Determine the integer values of \( n \) and \( m \)
For \( x = n\pi \):
- In the interval \([0, 20]\), \( n \) can take values \( 0, 1, 2, 3, 4, 5, 6\) (since \( 6\pi \approx 18.85 < 20 \)).
For \( x = \frac{m\pi}{2} \):
- In the interval \([0, 20]\), \( m \) can take values \( 0, 1, 2, \ldots, 12\) (since \( \frac{12\pi}{2} = 6\pi \approx 18.85 < 20 \)).
### Step 5: Identify the common solutions
The common solutions occur when:
- \( n\pi = \frac{m\pi}{2} \)
- This implies \( n = \frac{m}{2} \), meaning \( m \) must be even.
Let \( m = 2k \) where \( k \) is an integer. Then:
\[
x = \frac{2k\pi}{2} = k\pi
\]
Thus, \( k \) can take values \( 0, 1, 2, 3, 4, 5, 6\) (which corresponds to \( n = 0, 1, 2, 3, 4, 5, 6\)).
### Step 6: Count the solutions
The values of \( k \) that satisfy \( k\pi \) in the interval \([0, 20]\) are:
- \( k = 0 \) (0)
- \( k = 1 \) (π)
- \( k = 2 \) (2π)
- \( k = 3 \) (3π)
- \( k = 4 \) (4π)
- \( k = 5 \) (5π)
- \( k = 6 \) (6π)
Thus, there are a total of **7 solutions**.
### Final Answer
The number of solutions of \( \cos^2 x + \cos^2 2x = 2 \) in the interval \([0, 20]\) is **7**.
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