The number of solutions of the equation `(3+cos x)^(2)=4-2sin^(8)x" in "[0, 9pi)` is equal to

Options

A. 4
B. 5
C. 6
D. 7

Answer

To solve the equation \((3 + \cos x)^2 = 4 - 2 \sin^8 x\) in the interval \([0, 9\pi)\), we will follow these steps:

### Step 1: Analyze the Left-Hand Side (LHS)
The LHS is given by \((3 + \cos x)^2\). Since \(\cos x\) varies between \(-1\) and \(1\), we can find the range of \(3 + \cos x\):
- Minimum: \(3 + (-1) = 2\)
- Maximum: \(3 + 1 = 4\)

Thus, the range of \(3 + \cos x\) is \([2, 4]\). Squaring this gives:
\[ 4 \leq (3 + \cos x)^2 \leq 16 \]

### Step 2: Analyze the Right-Hand Side (RHS)
The RHS is \(4 - 2 \sin^8 x\). Since \(\sin x\) also varies between \(-1\) and \(1\), we have:
- Minimum of \(\sin^8 x\) is \(0\) (when \(\sin x = 0\))
- Maximum of \(\sin^8 x\) is \(1\) (when \(\sin x = \pm 1\))

Thus, the range of \(4 - 2 \sin^8 x\) is:
- Maximum: \(4 - 2(0) = 4\)
- Minimum: \(4 - 2(1) = 2\)

Therefore, the range of \(4 - 2 \sin^8 x\) is \([2, 4]\).

### Step 3: Set LHS equal to RHS
Since both sides of the equation are bounded between \(2\) and \(4\), we can equate them:
\[ (3 + \cos x)^2 = 4 - 2 \sin^8 x \]
This implies both sides can only equal \(4\) since that is the only value they share in their ranges.

### Step 4: Solve for \(\cos x\)
Setting the LHS equal to \(4\):
\[ (3 + \cos x)^2 = 4 \]
Taking the square root gives us two cases:
1. \(3 + \cos x = 2\)
2. \(3 + \cos x = -2\)

**Case 1:**
\[ 3 + \cos x = 2 \implies \cos x = -1 \]
This occurs at:
\[ x = (2n + 1)\pi \quad \text{for integers } n \]
In the interval \([0, 9\pi)\), the solutions are:
- \(x = \pi\)
- \(x = 3\pi\)
- \(x = 5\pi\)
- \(x = 7\pi\)

**Case 2:**
\[ 3 + \cos x = -2 \implies \cos x = -5 \]
This is not possible since \(\cos x\) must be in the range \([-1, 1]\).

### Step 5: Solve for \(\sin x\)
Setting the RHS equal to \(4\):
\[ 4 - 2 \sin^8 x = 4 \implies -2 \sin^8 x = 0 \implies \sin^8 x = 0 \]
This implies:
\[ \sin x = 0 \]
This occurs at:
\[ x = n\pi \quad \text{for integers } n \]
In the interval \([0, 9\pi)\), the solutions are:
- \(x = 0\)
- \(x = \pi\)
- \(x = 2\pi\)
- \(x = 3\pi\)
- \(x = 4\pi\)
- \(x = 5\pi\)
- \(x = 6\pi\)
- \(x = 7\pi\)
- \(x = 8\pi\)

### Step 6: Combine Solutions
From the two cases, we find:
- From \(\cos x = -1\): \(x = \pi, 3\pi, 5\pi, 7\pi\)
- From \(\sin x = 0\): \(x = 0, \pi, 2\pi, 3\pi, 4\pi, 5\pi, 6\pi, 7\pi, 8\pi\)

The unique solutions in the interval \([0, 9\pi)\) are:
- \(0, \pi, 2\pi, 3\pi, 4\pi, 5\pi, 6\pi, 7\pi, 8\pi\)

### Conclusion
The total number of unique solutions is \(9\).

### Final Answer
The number of solutions of the equation \((3 + \cos x)^2 = 4 - 2 \sin^8 x\) in the interval \([0, 9\pi)\) is \(9\).