Solve the Initial-Value Problem $( 1 + \cos x ) \frac { d y } { d x } = \left( 2 + e ^ { - y } \right) \sin x , y ( 0 ) = 0$

Solve the initial-value problem. $( 1 + \cos x ) \frac { d y } { d x } = \left( 2 + e ^ { - y } \right) \sin x , y ( 0 ) = 0$

A) $y = \ln ( 2 - \cos x ) ( 2 + \cos x )$
B) $y = \frac { ( 7 - \cos x ) } { ( 2 + \cos x ) }$
C) $y = \ln \frac { \cos x } { ( 2 + \cos x ) }$
D) $y = \frac { \ln ( 2 + \cos x ) } { \ln ( 2 - \cos x ) }$
E) $y = \ln ( 7 - \cos x ) - \ln ( 1 + \cos x )$

Correct Answer:

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