Помогите с решением, пожалуйста

$cos^2x \geq \frac{3sin^2x-1}{tg(\pi +x)} \; ,\; \; tgx\ \textgreater \ 0\\\\cos^2x \geq \frac{3sin^2x-1}{tgx} \; |\cdot tgx\ \textgreater \ 0\\\\tgx\cdot cos^2x \geq 3sin^2x-1\\\\ \frac{sinx}{cosx} \cdot cos^2x \geq 3sin^2x-1\\\\sinx\cdot cosx \geq 3sin^2x-(sin^2x+cos^2x)\\\\2sin^2x-sinx\cdot cosx-cos^2x \leq 0\; |cos^2x\ne 0\\\\2tg^2x-tgx-1 \leq 0\\\\D=9,\; \; (tgx)_1= \frac{1-3}{4}= -\frac{1}{2}\; ,\; (tgx)_2= \frac{1+3}{4} =1\\\\2t^2-t-1 \leq 0\; ,\; \; +++[-\frac{1}{2}\, ]---[\, 1\, ]+++$

$-\frac{1}{2} \leq t \leq 1$ $\Rightarrow \; \; \; -\frac{1}{2} \leq tgx \leq 1$

$tgx\ \textgreater \ 0\; \; \; \Rightarrow \; \; 0\ \textless \ tgx \leq 1\; \; \; \Rightarrow\\\\\pi n\ \textless \ x \leq \frac{\pi}{4}+\pi n,\; n\in Z\\\\x\in (\pi n\, ;\, \frac{\pi}{4}+\pi n\, ]\; ,\; n\in Z$