Solve for \( x \) : \( \sinh ^{2}\left(\frac{x}{2}\right)-\cosh x=-2 \)

Let's start by rewriting the given equation: \[ \sinh^2\left(\frac{x}{2}\right) - \cosh x = -2 \] We can use the identity that relates \(\cosh\) and \(\sinh\): \[ \cosh x = 2\sinh^2\left(\frac{x}{2}\right) + 1 \] Substituting this identity into the equation gives: \[ \sinh^2\left(\frac{x}{2}\right) - (2\sinh^2\left(\frac{x}{2}\right) + 1) = -2 \] This simplifies to: \[ \sinh^2\left(\frac{x}{2}\right) - 2\sinh^2\left(\frac{x}{2}\right) - 1 = -2 \] Combining like terms yields: \[ -\sinh^2\left(\frac{x}{2}\right) - 1 = -2 \] By adding 1 to both sides, we find: \[ -\sinh^2\left(\frac{x}{2}\right) = -1 \] This rearranges to: \[ \sinh^2\left(\frac{x}{2}\right) = 1 \] Taking the square root of both sides, we see that: \[ \sinh\left(\frac{x}{2}\right) = 1 \quad \text{or} \quad \sinh\left(\frac{x}{2}\right) = -1 \] Taking the inverse sinh on both scenarios gives: 1. \(\frac{x}{2} = \sinh^{-1}(1)\) leads to \(\frac{x}{2} = \ln(1 + \sqrt{2})\), hence \(x = 2\ln(1 + \sqrt{2})\). 2. \(\frac{x}{2} = \sinh^{-1}(-1)\) leads to \(\frac{x}{2} = -\ln(1 + \sqrt{2})\), hence \(x = -2\ln(1 + \sqrt{2})\). Putting it all together, we have the solutions: \[ x = 2\ln(1 + \sqrt{2}) \quad \text{and} \quad x = -2\ln(1 + \sqrt{2}) \]