We are tasked with solving the equation:

\[
\sin^{-1} x = \cot^{-1} (3x)
\]

### Step-by-step Solution:

1. **Start with the basic definitions of inverse trigonometric functions:**
   - \(\sin^{-1} x\) is the angle whose sine is \(x\). Let's assume this angle is \(\theta\), so:
     \[
     \sin^{-1} x = \theta \quad \Rightarrow \quad \sin \theta = x
     \]

   - \(\cot^{-1} (3x)\) is the angle whose cotangent is \(3x\). Let's assume this angle is \(\phi\), so:
     \[
     \cot^{-1} (3x) = \phi \quad \Rightarrow \quad \cot \phi = 3x
     \]

   Thus, the equation becomes:
   \[
   \theta = \phi
   \]
   And we now have two relationships:
   \[
   \sin \theta = x \quad \text{and} \quad \cot \theta = 3x
   \]

2. **Express \(\cot \theta\) in terms of \(\sin \theta\):**
   We know that:
   \[
   \cot \theta = \frac{\cos \theta}{\sin \theta}
   \]
   Substituting the second equation \(\cot \theta = 3x\), we get:
   \[
   \frac{\cos \theta}{\sin \theta} = 3x
   \]

3. **Use the identity \(\sin^2 \theta + \cos^2 \theta = 1\):**
   From \(\sin \theta = x\), we know that:
   \[
   \cos^2 \theta = 1 - \sin^2 \theta = 1 - x^2
   \]
   So, \(\cos \theta = \sqrt{1 - x^2}\) (since \(\theta\) is the angle of the inverse sine function, we restrict the value to the positive root).

4. **Substitute \(\cos \theta\) in the equation:**
   We now have:
   \[
   \frac{\sqrt{1 - x^2}}{x} = 3x
   \]

5. **Solve for \(x\):**
   Multiply both sides by \(x\) (assuming \(x \neq 0\)):
   \[
   \sqrt{1 - x^2} = 3x^2
   \]
   Square both sides:
   \[
   1 - x^2 = 9x^4
   \]
   Rearrange the terms:
   \[
   9x^4 + x^2 - 1 = 0
   \]
   This is a quadratic equation in terms of \(x^2\). Let \(y = x^2\), so the equation becomes:
   \[
   9y^2 + y - 1 = 0
   \]

6. **Solve the quadratic equation:**
   Using the quadratic formula:
   \[
   y = \frac{-1 \pm \sqrt{1^2 - 4(9)(-1)}}{2(9)}
   \]
   Simplifying:
   \[
   y = \frac{-1 \pm \sqrt{1 + 36}}{18} = \frac{-1 \pm \sqrt{37}}{18}
   \]
   So the two possible solutions for \(y\) are:
   \[
   y = \frac{-1 + \sqrt{37}}{18} \quad \text{or} \quad y = \frac{-1 - \sqrt{37}}{18}
   \]
   Since \(y = x^2\) must be non-negative, we discard the negative root and keep:
   \[
   y = \frac{-1 + \sqrt{37}}{18}
   \]
   Therefore:
   \[
   x^2 = \frac{-1 + \sqrt{37}}{18}
   \]
   Taking the square root:
   \[
   x = \pm \sqrt{\frac{-1 + \sqrt{37}}{18}}
   \]

Thus, the solutions for \(x\) are \(x = \pm \sqrt{\frac{-1 + \sqrt{37}}{18}}\).

---

Would you like more details on any step? Here are 5 related questions to explore:

1. What are the properties of inverse trigonometric functions?
2. How does squaring both sides affect the solution of trigonometric equations?
3. What are common strategies for solving quadratic equations?
4. What conditions determine whether you should take the positive or negative square root in trigonometric solutions?
5. How can we graphically interpret the solutions to \(\sin^{-1} x = \cot^{-1} (3x)\)?

**Tip**: When dealing with inverse trigonometric functions, always pay attention to the principal range of the functions to avoid extraneous solutions.

Explore a detailed solution to the equation sin^-1 x = cot^-1 (3x). This problem demonstrates concepts in inverse trigonometric functions, algebraic manipulation, and quadratic equations. Perfect for students in Grades 11-12 interested in mastering advanced trigonometric equations.

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