Ответ:

[tex]y'tgx = 1 + y \\ \frac{dy}{dx}= \frac{1 + y}{tgx} \\ \int \frac{dy}{1 + y} = \int \frac{dx}{tgx} \\ [/tex]

.....

[tex] \int \frac{dx}{tgx} = \int \: ctgxdx = \int \frac{cosx}{sinx} dx = \\ = |t = sinx ;dt = cosxdx| = \\ = \int \frac{dt}{t} = ln |t| = ln |sinx| + lnС[/tex]

.....

[tex]ln |y + 1| = ln |sinx| + lnС \\ y = Сsinx - 1[/tex]

[tex]y = - \frac{1}{2} ;x = \frac{\pi}{6} [/tex]

[tex] - \frac{1}{2} = Сsin \frac{\pi}{6} - 1 \\ \frac{1}{2} = \frac{1}{2} С \\ С = 1[/tex]

[tex]y = sinx - 1[/tex]