What is the general solution for ((1-tanh(2x)))/((1+tanh(2x)))=2 ?

The general solution for ((1-tanh(2x)))/((1+tanh(2x)))=2 is x=-1/4 ln(2)

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((1-tanh(2x)))/((1+tanh(2x)))=2

\frac{( 1 - tanh ( 2 x ) )}{( 1 + tanh ( 2 x ) )} = 2

x = - \frac{1}{4} ln (2)

Solution steps

\frac{( 1 - tanh ( 2 x ) )}{( 1 + tanh ( 2 x ) )} = 2

Rewrite using trig identities

\frac{1 - \frac{e ^{2 x} - e ^{- 2 x}}{e ^{2 x} + e ^{- 2 x}}}{1 + \frac{e ^{2 x} - e ^{- 2 x}}{e ^{2 x} + e ^{- 2 x}}} = 2

\frac{1 - \frac{e ^{2 x} - e ^{- 2 x}}{e ^{2 x} + e ^{- 2 x}}}{1 + \frac{e ^{2 x} - e ^{- 2 x}}{e ^{2 x} + e ^{- 2 x}}} = 2 : x = - \frac{1}{4} ln (2)

x = - \frac{1}{4} ln (2)

sin (4 θ) = 0.345

- 3 = cot^{2} (x) + 3 csc (x)

4 cos^{2} (θ) + 3 cos (θ) = 1

x, 2 sin^{2} (x) + cos (x) = 1

cos^{2} (x) - 1 = 1 - 2 sin^{2} (x)

What is the general solution for ((1-tanh(2x)))/((1+tanh(2x)))=2 ?
The general solution for ((1-tanh(2x)))/((1+tanh(2x)))=2 is x=-1/4 ln(2)