What is the solution for (dy)/(dx)=-1/(sin(y)),y(-3)= pi/2 ?

The solution for (dy)/(dx)=-1/(sin(y)),y(-3)= pi/2 is y=arccos(x+3)+2pin,y=-arccos(x+3)+2pin

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(dy)/(dx)=-1/(sin(y)),y(-3)= pi/2

\frac{d y}{d x} = - \frac{1}{sin ( y )}, y (- 3) = \frac{π}{2}

y = arccos (x + 3) + 2 πn, y = - arccos (x + 3) + 2 πn

Solution steps

\frac{d y}{d x} = - \frac{1}{sin ( y )}

Solve separable ODE:

y = arccos (x + 3) + 2 πn, y = - arccos (x + 3) + 2 πn

y = arccos (x + 3) + 2 πn, y = - arccos (x + 3) + 2 πn

\frac{d y}{d x} (- y cot (x)) = \frac{sin ( 2 x )}{2}

y^{^{'}} = \frac{x ^{2} + 7 x + 3}{y ^{2}}, y (0) = 3

4 d y + y d x = x^{2} d y, y (4) = - 1

y^{^{'}} + (x + 2) y = 0

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{7} e^{- x} d y = 0

What is the solution for (dy)/(dx)=-1/(sin(y)),y(-3)= pi/2 ?
The solution for (dy)/(dx)=-1/(sin(y)),y(-3)= pi/2 is y=arccos(x+3)+2pin,y=-arccos(x+3)+2pin