Sinx+tgx=1/cosx-cosx

ОДЗ:
$cosx \neq 0 \\ \\ x \neq \dfrac{ \pi }{2} + \pi m, \ m \in Z$

$sinx + tgx = \dfrac{1}{cosx} - cosx \\ \\ sinx + \dfrac{sinx}{cosx} = \dfrac{1}{cosx} - cosx \\ \\ \dfrac{sinxcosx}{cosx} + \dfrac{sinx}{cosx} = \dfrac{1}{cosx} - \dfrac{cos^2x }{cosx} \\ \\ \dfrac{sinxcosx + sinx}{cosx} = \dfrac{1 - cos^2x}{cosx} \\ \\ sinxcosx + sinx = sin^2x \\ \\ sin^2x - sinxcosx - sinx = 0 \\ \\ sinx(sinx - cosx - 1) = 0 \\ \\ sinx = 0 \\ \\ \boxed{x = \pi n, \ n \in Z }$

$sinx - cosx - 1 = 0 \\ \\ sinx - cosx = 1 \\ \\ \dfrac{ \sqrt{2} }{2} sinx - \dfrac{ \sqrt{2} }{2} cosx = 1 \\ \\ sinx \cdot cos \dfrac{ \pi }{4} - cosx \cdot sin \dfrac{ \pi }{4} = 1 \\ \\ sinx \bigg (x - \dfrac{ \pi }{4} \bigg ) = 1 \\ \\ x - \dfrac{ \pi }{4} = \dfrac{ \pi }{2} + 2 \pi k, \ k \in Z \\ \\ x = \dfrac{ \pi} {4} + \dfrac{ \pi }{2} + 2 \pi k, \ k \in Z \\ \\ \boxed{x = \dfrac{ 3\pi} {4} + 2 \pi k, \ k \in Z}$

Ответ: $\boxed{x = \pi n, \ n \in Z ; \ \dfrac{ 3\pi} {4} + 2 \pi k, \ k \in Z. }$