cschx,coshx and tanhx

Here's the first one:

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\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}=75\)

1. Given that cschx=-9/40, find the exact value of coshx and tanh2x

\(\begin{array}{rcll} csch(x) & =& -\frac{9}{40} \\\\ cosh(x) & =& \frac{ \sqrt{1+csch^2(x)} } { csch(x) }\\ cosh(x) & =& \frac{ \sqrt{1+ (-\frac{9}{40})^2 } } { -\frac{9}{40} }\\ cosh(x) & =& -\frac{40}{9} \cdot \sqrt{1+ (-\frac{9}{40})^2 } \\ cosh(x) & =& -\frac{40}{9} \cdot \sqrt{1+ \frac{81}{1600}) } \\ cosh(x) & =& -\frac{40}{9} \cdot \sqrt{ \frac{1681}{1600}) } \\ cosh(x) & =& -\frac{40}{9} \cdot \frac{41}{40} \\\\ cosh(x) & =& -\frac{41}{9} \\ \hline \\ sinh(x) & =& \frac{ 1 } { csch(x) }\\ sinh(x) & =& \frac{ 1 } { -\frac{9}{40} }\\ sinh(x) & =& -\frac{40}{9}\\ \hline \\ \end{array} \\\)

\(\begin{array}{rcll} sinh (2x) &=& 2\cdot sinh (x)\cdot cosh (x)\\ cosh (2x) &=& cosh^2(x) + sinh^2(x) = 2 \cdot cosh^2(x) — 1 = 1 + 2\cdot sinh^2(x)\\\\ tanh (2x) &=& \frac{ sinh(2x) } { cosh(2x) } \\ tanh (2x) &=& \frac{ 2\cdot sinh (x)\cdot cosh (x) } { 1 + 2\cdot sinh^2(x) } \\ tanh (2x) &=& \frac{ 2\cdot (-\frac{40}{9})\cdot (-\frac{41}{9}) } { 1 + 2\cdot (-\frac{40}{9})^2 } \\ tanh (2x) &=& \frac{ 2\cdot (\frac{40\cdot 41}{9\cdot 9}) } { 1 + 2\cdot (\frac{40^2}{9^2}) } \\ tanh (2x) &=& \frac{ 2\cdot (\frac{40\cdot 41}{9^2}) } { \frac{9^2 +2\cdot 40^2}{9^2} } \\ tanh (2x) &=& \frac{ 2\cdot 40 \cdot 41 } { 9^2 +2\cdot 40^2 } \\\\ tanh (2x) &=& \frac{ 3280 } { 3281 }\\ \hline \\ \end{array}\)

2. Solve the equation

\(x=tanh(~ \ln{(~ \sqrt{6x}~) } ~) \quad \text{ for } 0 < x < 1 \)

\(\begin{array}{rcll} x &=& tanh(~ \ln{(~ \sqrt{6x}~) } ~) \qquad & | \qquad tanh^{-1}() \\ tanh^{-1}(x) &=& \ln{(~ \sqrt{6x}~) } \qquad & | \qquad tanh^{-1}(x) = \frac12\cdot \ln{(~\frac{1+x}{1-x}~)} \quad \text{ for } \quad -1<x<1\\ \frac12\cdot \ln{(~\frac{1+x}{1-x}~)} &=& \ln{(~ \sqrt{6x}~) } \\ \ln{(~[\frac{1+x}{1-x}]^\frac12~)} &=& \ln{(~ \sqrt{6x}~) } \\ \left[\frac{1+x}{1-x} \right]^\frac12 &=& \sqrt{6x}\\ \sqrt{\frac{1+x}{1-x} } &=& \sqrt{6x} \qquad & | \qquad (\text{square both sides})\\ \frac{1+x}{1-x} &=& 6x \\ 1+x &=& 6x\cdot ( 1-x ) \\ 1+x &=& 6x - 6x^2 \\ 6x^2 -6x + x + 1 &=& 0 \\ \end{array}\)

\(\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}\)

\(\small{ \begin{array}{rcll} 6x^2 -5x + 1 &=& 0 \qquad & a = 6 \qquad b = -5 \qquad c = 1\\ x_{1,2} &=& \dfrac{-(-5) \pm \sqrt{(-5)^2-4\cdot 6 \cdot 1} }{2\cdot 6} \\ x_{1,2} &=& \dfrac{ 5 \pm \sqrt{25 - 24} }{ 12 } \\ x_{1,2} &=& \dfrac{ 5 \pm \sqrt{1} }{ 12 } \\ x_{1,2} &=& \dfrac{ 5 \pm 1 }{ 12 } \\\\ x_1 &=& \dfrac{ 5 + 1 }{ 12 } \\ x_1 &=& \dfrac{ 6 }{ 12 } \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\dfrac{ 1 }{ 2 } }\\ \hline \\ x_2 &=& \dfrac{ 5 - 1 }{ 12 } \\ x_2 &=& \dfrac{ 4 }{ 12 } \\ \mathbf{x_2} &\mathbf{=}& \mathbf{\dfrac{ 1 }{ 3 } } \\ \hline \end{array} } \)

{nl}

Here's my version of the second one:

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