What is the general solution for cosh(z)=1,cosh(z)=-2 ?

The general solution for cosh(z)=1,cosh(z)=-2 is z=0

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cosh(z)=1,cosh(z)=-2

cosh (z) = 1, cosh (z) = - 2

z = 0

Solution steps

cosh (z) = 1, cosh (z) = - 2

Rewrite using trig identities

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

\frac{e ^{z} + e ^{- z}}{2} = 1 : z = 0

z = 0

tan (2 x) = \frac{- 3}{3}

x, - 4 cos (x) = - sin^{2} (x) + 4

5 cos (c) + 23 = 0

cos (2 θ) = - \frac{28}{53}

cos (θ) = \frac{( 1 )}{( 2 )}

What is the general solution for cosh(z)=1,cosh(z)=-2 ?
The general solution for cosh(z)=1,cosh(z)=-2 is z=0