It seems like your query got cut off. From what I can infer, you're asking for the solution of an equation that involves hyperbolic functions like $\sinh(x)$ and $\cosh(x)$, and you'd like to solve for $\tanh(x)$. Since the full equation wasn't provided, I'll assume it might look something like this:

$6\sinh(x) + 2\cosh(x) = 0$

Step-by-Step Solution:

1. Rewrite the equation: $6\sinh(x) + 2\cosh(x) = 0$ Divide the entire equation by 2: $3\sinh(x) + \cosh(x) = 0$

2. Express $\sinh(x)$ and $\cosh(x)$ in terms of exponentials: $\sinh(x) = \frac{e^x - e^{-x}}{2}, \quad \cosh(x) = \frac{e^x + e^{-x}}{2}$ Substituting these into the equation: $3\left(\frac{e^x - e^{-x}}{2}\right) + \frac{e^x + e^{-x}}{2} = 0$ Multiply everything by 2 to eliminate the denominators: $3(e^x - e^{-x}) + (e^x + e^{-x}) = 0$

3. Simplify the equation: Combine like terms: $3e^x - 3e^{-x} + e^x + e^{-x} = 0$ This simplifies to: $4e^x - 2e^{-x} = 0$

4. Multiply through by $e^x$ to eliminate the negative exponent: $4e^{2x} - 2 = 0$ Solve for $e^{2x}$: $4e^{2x} = 2 \quad \Rightarrow \quad e^{2x} = \frac{1}{2}$

5. Take the natural logarithm: $2x = \ln\left(\frac{1}{2}\right) = -\ln(2)$ Solve for $x$: $x = -\frac{\ln(2)}{2}$

6. Find $\tanh(x)$: Now, recall that: $\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$ Substituting $x = -\frac{\ln(2)}{2}$ into this expression for $\tanh(x)$, we get: $\tanh\left(-\frac{\ln(2)}{2}\right)$ Since the exact value can be computed, the result is $-\frac{1}{3}$.

Thus, $\tanh(x) = -\frac{1}{3}$.

---

Would you like further clarification or details on any step? Here are some related questions:

1. How do the hyperbolic functions $\sinh(x)$ and $\cosh(x)$ relate to exponential functions?
2. Can you explain why multiplying by $e^x$ eliminates the negative exponent?
3. How can we derive the natural logarithm from exponential equations?
4. What are some real-world applications of hyperbolic functions like $\tanh(x)$?
5. How is the solution process similar for other transcendental equations?

Tip: Hyperbolic functions are closely related to their trigonometric counterparts but involve exponential expressions.