pt bậc 1 đối với sinx cosx tanx+2cotx-3=0 tìm TXĐ của hàm số y=5tan^2x+4cot^2x y=2tanx+3/sin2x y=x+1/sin2x+3

\[\begin{array}{l} + )\,\,\,\tan x + 2\cot x - 3 = 0\\ DK:\,\,\,\left\{ \begin{array}{l} \sin x \ne 0\\ \cos x \ne 0 \end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne \frac{{k\pi }}{2}\\ \Rightarrow D = R\backslash \left\{ {\frac{{k\pi }}{2}} \right\}.\\ \Rightarrow pt \Leftrightarrow \tan x + 2.\frac{1}{{\tan x}} - 3 = 0\\ \Leftrightarrow {\tan ^2}x - 3\tan x + 2 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \tan x = 1\\ \tan x = 2 \end{array} \right. \Leftrightarrow ....\\ Tim\,\,\,TXD\,\,\,cua\,\,\,ham\,\,so:\\ + )\,\,\,y = 5{\tan ^2}x + 4{\cot ^2}x\\ DK:\,\,\,\left\{ \begin{array}{l} \cos x \ne 0\\ \sin x \ne 0 \end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne \frac{{k\pi }}{2}\\ \Rightarrow D = R\backslash \left\{ {\frac{{k\pi }}{2}} \right\}.\\ + )\,\,y = \frac{{2\tan x + 3}}{{\sin 2x}}\\ DK:\,\,\,\left\{ \begin{array}{l} \cos x \ne 0\\ \sin 2x \ne 0 \end{array} \right. \Leftrightarrow \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne \frac{{k\pi }}{2}\\ \Rightarrow D = R\backslash \left\{ {\frac{{k\pi }}{2}} \right\}.\\ + )\,\,\,y = \frac{{x + 1}}{{\sin 2x + 3}}\\ \sin 2x + 3 > 0\,\,\forall x\\ \Rightarrow D = R. \end{array}\]