Solve the initial-value problem.
$( 1 + \cos x ) \frac { d y } { d x } = \left( 6 + e ^ { - y } \right) \sin x , y ( 0 ) = 0$
A) $y = \frac { \ln ( 6 + \cos x ) } { \ln ( 6 - \cos x ) }$
B)
$y = \ln \left( \frac { 1 } { 6 } \left( \sec ^ { 12 } \left( \frac { x } { 2 } \right) - 1 \right) \right.$
C) $y = \ln ( 6 - \cos x ) ( 6 + \cos x )$
D)
$y = \ln \left( \frac { 1 } { 6 } \left( 7 \sec ^ { 12 } \left( \frac { x } { 2 } \right) - 1 \right) \right.$
E)
$y = \ln \left( 6 \sec \left( \frac { x } { 2 } \right) - 1 \right)$

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The Answer of Solve the initial-value problem.
$( 1 + \cos x ) \frac { d y } { d x } = \left( 6 + e ^ { - y } \right) \sin x , y ( 0 ) = 0$
A) $y = \frac { \ln ( 6 + \cos x ) } { \ln ( 6 - \cos x ) }$
B)
$y = \ln \left( \frac { 1 } { 6 } \left( \sec ^ { 12 } \left( \frac { x } { 2 } \right) - 1 \right) \right.$
C) $y = \ln ( 6 - \cos x ) ( 6 + \cos x )$
D)
$y = \ln \left( \frac { 1 } { 6 } \left( 7 \sec ^ { 12 } \left( \frac { x } { 2 } \right) - 1 \right) \right.$
E)
$y = \ln \left( 6 \sec \left( \frac { x } { 2 } \right) - 1 \right)$

Solve the Initial-Value Problem (1+cos⁡x)dydx=(6+e−y)sin⁡x,y(0)=0( 1 + \cos x ) \frac { d y } { d x } = \left( 6 + e ^ { - y } \right) \sin x , y ( 0 ) = 0(1+cosx)dxdy​=(6+e−y)sinx,y(0)=0

Solve the Initial-Value Problem $( 1 + \cos x ) \frac { d y } { d x } = \left( 6 + e ^ { - y } \right) \sin x , y ( 0 ) = 0$