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You asked:
Solve the equation
cosh ( 47.98 L ) = − 0.0030650 \cosh\left( 47.98 L \right) = -0.0030650 cosh ( 47.98 L ) = − 0.0030650 for the variable
L L L .
MathBot Answer:
The complex solutions are: L = { 50 i ( − π + 2 π n + arctan ( 39999624231 613 ) ) 2399 | n ∈ Z } ≈ { 0.020842018 i ( 6.2831853 n − 1.5738613 ) | n ∈ Z } L = { 50 i ( π − arctan ( 39999624231 613 ) + 2 π n ) 2399 | n ∈ Z } ≈ { 0.020842018 i ( 6.2831853 n + 1.5738613 ) | n ∈ Z } \begin{aligned}L &= \left\{\frac{50 i \left(- \pi + 2 \pi n + \arctan{\left(\frac{\sqrt{39999624231}}{613} \right)}\right)}{2399}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{0.020842018 i \left(6.2831853 n - 1.5738613\right)\; \middle|\; n \in \mathbb{Z}\right\}\\L &= \left\{\frac{50 i \left(\pi - \arctan{\left(\frac{\sqrt{39999624231}}{613} \right)} + 2 \pi n\right)}{2399}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{0.020842018 i \left(6.2831853 n + 1.5738613\right)\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned} L L = ⎩ ⎨ ⎧ 2399 50 i ( − π + 2 πn + arctan ( 613 39999624231 ) ) n ∈ Z ⎭ ⎬ ⎫ ≈ { 0.020842018 i ( 6.2831853 n − 1.5738613 ) ∣ n ∈ Z } = ⎩ ⎨ ⎧ 2399 50 i ( π − arctan ( 613 39999624231 ) + 2 πn ) n ∈ Z ⎭ ⎬ ⎫ ≈ { 0.020842018 i ( 6.2831853 n + 1.5738613 ) ∣ n ∈ Z }
i i i imaginary unit , defined as i 2 = − 1 i^2 = -1 i 2 = − 1
