The question you uploaded asks to:

(a) Express $\sinh(2\theta)$ and $\cosh(2\theta)$ in terms of $\tanh(\theta)$, and then solve the equation: $2\cosh(2\theta) + \sinh(2\theta) = 2$ Leave the answer in logarithmic form.

Step 1: Express $\sinh(2\theta)$ and $\cosh(2\theta)$ in terms of $\tanh(\theta)$

Using the double angle identities for hyperbolic functions:

$\sinh(2\theta) = 2\sinh(\theta)\cosh(\theta)$ $\cosh(2\theta) = \cosh^2(\theta) + \sinh^2(\theta)$

Now, recall that: $\tanh(\theta) = \frac{\sinh(\theta)}{\cosh(\theta)}$ Thus, $\sinh(\theta) = \tanh(\theta)\cosh(\theta)$

We can substitute this back into the double angle formulas.

Step 2: Simplify $\sinh(2\theta)$ and $\cosh(2\theta)$

1. $\sinh(2\theta)$: $\sinh(2\theta) = 2 \sinh(\theta)\cosh(\theta) = 2 \tanh(\theta)\cosh^2(\theta)$ Since $\cosh^2(\theta) = \frac{1}{1 - \tanh^2(\theta)}$, we substitute: $\sinh(2\theta) = 2 \frac{\tanh(\theta)}{1 - \tanh^2(\theta)}$

2. $\cosh(2\theta)$: $\cosh(2\theta) = \cosh^2(\theta) + \sinh^2(\theta) = \frac{1}{1 - \tanh^2(\theta)}$

Step 3: Solve the equation

Now that we have $\sinh(2\theta)$ and $\cosh(2\theta)$ in terms of $\tanh(\theta)$, we substitute these into the equation:

$2 \cosh(2\theta) + \sinh(2\theta) = 2$ Substitute $\cosh(2\theta) = \frac{1}{1 - \tanh^2(\theta)}$ and $\sinh(2\theta) = 2 \frac{\tanh(\theta)}{1 - \tanh^2(\theta)}$: $2 \left( \frac{1}{1 - \tanh^2(\theta)} \right) + 2 \frac{\tanh(\theta)}{1 - \tanh^2(\theta)} = 2$

Factor out the common denominator: $\frac{2 + 2\tanh(\theta)}{1 - \tanh^2(\theta)} = 2$

Multiply both sides by $1 - \tanh^2(\theta)$: $2 + 2\tanh(\theta) = 2 (1 - \tanh^2(\theta))$

Expand the right-hand side: $2 + 2\tanh(\theta) = 2 - 2\tanh^2(\theta)$

Cancel the constant terms on both sides: $2\tanh(\theta) = -2\tanh^2(\theta)$

Divide both sides by 2: $\tanh(\theta) = -\tanh^2(\theta)$

Rearrange the equation: $\tanh^2(\theta) + \tanh(\theta) = 0$

Factor the equation: $\tanh(\theta)(\tanh(\theta) + 1) = 0$

Thus, $\tanh(\theta) = 0$ or $\tanh(\theta) = -1$.

Step 4: Solve for $\theta$ in logarithmic form

1. If $\tanh(\theta) = 0$, then $\theta = 0$.

2. If $\tanh(\theta) = -1$, we use the definition of $\tanh(\theta)$: $\tanh(\theta) = \frac{e^{2\theta} - 1}{e^{2\theta} + 1} = -1$ Solving this equation: $\frac{e^{2\theta} - 1}{e^{2\theta} + 1} = -1$ Cross-multiply: $e^{2\theta} - 1 = - (e^{2\theta} + 1)$ $e^{2\theta} - 1 = -e^{2\theta} - 1$ Simplify: $2e^{2\theta} = 0$ Thus, $\theta = 0$.

Therefore, the solution is $\theta = 0$.

Do you need more details or have any questions?

Here are five related questions for further practice:

1. Can you derive the identity for $\tanh(2\theta)$ in terms of $\tanh(\theta)$?
2. How do the graphs of $\sinh(x)$ and $\cosh(x)$ compare?
3. Solve $\cosh(2\theta) - \sinh(2\theta) = 1$ for $\theta$.
4. How would you express $\tanh^{-1}(x)$ in terms of logarithms?
5. What is the relationship between the hyperbolic and trigonometric functions?

Tip: Always keep the identities of hyperbolic functions handy to simplify complex equations!