ПОМОГИТЕ,ОЧЕНЬ НАДО!!!! (корень из 1+sinx )+cosx= 0

[latex] \sqrt{1+sin(x)} = - cos(x)[/latex] Возведем в квадрат: [latex] (\sqrt{1+sin(x)})^{2} = (-cos(x))^2 \\ 1+ sin(x) - cos^2(x) = 0 \\ sin^2(x) + sin(x) = 0 \\ sin(x)*(sin(x)+1) = 0 \\ sin(x) = 0; x = 0 + n \pi \\ sin(x) = -1; x = \frac{3 \pi}{2} + 2 \pi n [/latex]