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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!

Express sinh(2θ) and cosh(2θ) in terms of tanh(θ) and hence solve the equation 2cosh(2θ) + sinh(2θ) = 2, leave your answer in logarithmic form.

Step-by-step solution for expressing sinh(2θ) and cosh(2θ) in terms of tanh(θ) and solving the equation 2cosh(2θ) + sinh(2θ) = 2. The solution includes algebraic manipulation and leaves the answer in logarithmic form. Suitable for advanced algebra students.

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