What is the general solution for cosh(θ)=4-j ?

The general solution for cosh(θ)=4-j is θ=ln(-j+4+sqrt(j^2-8j+15)),θ=ln(-j+4-sqrt(j^2-8j+15))

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cosh(θ)=4-j

cosh (θ) = 4 - j

θ = ln (- j + 4 + j^{2} - 8 j + 15), θ = ln (- j + 4 - j^{2} - 8 j + 15)

Solution steps

cosh (θ) = 4 - j

Rewrite using trig identities

\frac{e ^{θ} + e ^{- θ}}{2} = 4 - j

\frac{e ^{θ} + e ^{- θ}}{2} = 4 - j : θ = ln (- j + 4 + j^{2} - 8 j + 15), θ = ln (- j + 4 - j^{2} - 8 j + 15)

θ = ln (- j + 4 + j^{2} - 8 j + 15), θ = ln (- j + 4 - j^{2} - 8 j + 15)

\frac{1}{2} = tan (x)

7 sec (θ) = - 9.2

cos (x) = \frac{3 3}{2 π}

\frac{sin ( b )}{19} = \frac{sin ( 7 4 ^{\circ} )}{19}

- 8 cos^{2} (x) + 6 = 2

What is the general solution for cosh(θ)=4-j ?
The general solution for cosh(θ)=4-j is θ=ln(-j+4+sqrt(j^2-8j+15)),θ=ln(-j+4-sqrt(j^2-8j+15))