Find number of solutions of the equation `sin^(-1)(|log_(6)^(2)(cos x)-1|)+cos^(-1)(|3log_(6)^(2)(cos x)-7|)=(pi)/(2)`, if `x in [0, 4pi]`.

Answer

To solve the equation

\[ \sin^{-1}(|\log_6^2(\cos x) - 1|) + \cos^{-1}(|3\log_6^2(\cos x) - 7|) = \frac{\pi}{2} \]

for \( x \in [0, 4\pi] \), we can follow these steps:

### Step 1: Set up the substitution
Let \( t = \log_6(\cos x) \). Then the equation becomes:

\[ \sin^{-1}(|t^2 - 1|) + \cos^{-1}(|3t^2 - 7|) = \frac{\pi}{2} \]

### Step 2: Use the identity
Using the identity \( \sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2} \), we can conclude that:

\[ |t^2 - 1| = |3t^2 - 7| \]

### Step 3: Analyze the cases for absolute values
We need to consider the cases for the absolute values.

#### Case 1: \( t^2 - 1 = 3t^2 - 7 \)
This simplifies to:

\[ -2t^2 + 6 = 0 \implies t^2 = 3 \implies t = \pm \sqrt{3} \]

#### Case 2: \( t^2 - 1 = -(3t^2 - 7) \)
This simplifies to:

\[ t^2 - 1 = -3t^2 + 7 \implies 4t^2 - 8 = 0 \implies t^2 = 2 \implies t = \pm \sqrt{2} \]

#### Case 3: \( -t^2 + 1 = 3t^2 - 7 \)
This simplifies to:

\[ 4t^2 - 8 = 0 \implies t^2 = 2 \implies t = \pm \sqrt{2} \]

#### Case 4: \( -t^2 + 1 = -(3t^2 - 7) \)
This simplifies to:

\[ -t^2 + 1 = -3t^2 + 7 \implies 2t^2 - 6 = 0 \implies t^2 = 3 \implies t = \pm \sqrt{3} \]

### Step 4: Collect all solutions
From the cases, we have the solutions:

1. \( t = \sqrt{3} \)
2. \( t = -\sqrt{3} \)
3. \( t = \sqrt{2} \)
4. \( t = -\sqrt{2} \)

### Step 5: Convert back to \( x \)
Recall \( t = \log_6(\cos x) \), thus:

\[ \cos x = 6^t \]

For each \( t \):

1. For \( t = \sqrt{3} \): \( \cos x = 6^{\sqrt{3}} \) (not valid since \( \cos x \) must be in \([-1, 1]\))
2. For \( t = -\sqrt{3} \): \( \cos x = 6^{-\sqrt{3}} \) (valid)
3. For \( t = \sqrt{2} \): \( \cos x = 6^{\sqrt{2}} \) (not valid)
4. For \( t = -\sqrt{2} \): \( \cos x = 6^{-\sqrt{2}} \) (valid)

### Step 6: Find the values of \( x \)
Now we need to find the values of \( x \) for:

- \( \cos x = 6^{-\sqrt{3}} \)
- \( \cos x = 6^{-\sqrt{2}} \)

Since both values are positive and less than 1, we can find \( x \) in the interval \( [0, 4\pi] \):

For each valid \( \cos x \):

1. \( \cos x = 6^{-\sqrt{3}} \) gives two solutions in \( [0, 4\pi] \)
2. \( \cos x = 6^{-\sqrt{2}} \) gives another two solutions in \( [0, 4\pi] \)

### Conclusion
Thus, the total number of solutions is:

\[ \text{Total Solutions} = 2 + 2 = 4 \]

### Final Answer
The number of solutions of the equation is **4**.