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 find the number of solutions for 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: Understanding the Equation
Recall that \( \sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2} \) for any \( a \) in the range \([-1, 1]\). This implies that:

\[ |\log_6^2(\cos x) - 1| = |3\log_6^2(\cos x) - 7| \]

### Step 2: Setting Up the Equation
We can set \( t = \log_6^2(\cos x) \). The equation simplifies to:

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

### Step 3: Solving the Absolute Value Equation
We need to consider two cases based on the properties of absolute values.

**Case 1:** \( t - 1 = 3t - 7 \)

\[ t - 1 = 3t - 7 \implies 2t = 6 \implies t = 3 \]

**Case 2:** \( t - 1 = -(3t - 7) \)

\[ t - 1 = -3t + 7 \implies 4t = 8 \implies t = 2 \]

**Case 3:** \( -(t - 1) = 3t - 7 \)

\[ - t + 1 = 3t - 7 \implies 4t = 8 \implies t = 2 \]

**Case 4:** \( -(t - 1) = -(3t - 7) \)

\[ - t + 1 = -3t + 7 \implies 2t = 6 \implies t = 3 \]

### Step 4: Summary of Solutions
From the above cases, we have two unique solutions for \( t \):

1. \( t = 2 \)
2. \( t = 3 \)

### Step 5: Relating \( t \) Back to \( x \)
Recall that \( t = \log_6^2(\cos x) \). Thus, we have:

1. For \( t = 2 \):
\[ \log_6^2(\cos x) = 2 \implies \cos x = 6^{\pm \sqrt{2}} \text{ (which is outside the range of cosine)} \]

2. For \( t = 3 \):
\[ \log_6^2(\cos x) = 3 \implies \cos x = 6^{\pm \sqrt{3}} \text{ (which is also outside the range of cosine)} \]

### Step 6: Conclusion
Since both values of \( t \) lead to values of \( \cos x \) that are not in the range \([-1, 1]\), there are no valid solutions for \( x \) in the interval \([0, 4\pi]\).

Thus, the number of solutions is:

\[ \boxed{0} \]