Đáp án:
$1)D=\mathbb{R} \setminus \left\{ \dfrac{\pi}{2}+k \pi;k \in \mathbb{Z}\right\}\\ 2)\\ a)\left[\begin{array}{l} x =\dfrac{\pi}{6}+k 2 \pi (k \in \mathbb{Z}) \\ x =\dfrac{5\pi}{6}+k 2 \pi (k \in \mathbb{Z})\end{array} \right.\\ b) x=\pm \dfrac{3\pi}{4}+k 2 \pi (k \in \mathbb{Z})\\ c)x=\pi+k 2 \pi (k \in \mathbb{Z})\\ d) x=k 2 \pi (k \in \mathbb{Z}).$
Giải thích các bước giải:
$1)\\ y = \tan^2x=\dfrac{\sin^2x}{\cos^2x}\\ \text{ĐKXĐ: } \cos^2x \ne 0 \Leftrightarrow \cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{2}+k \pi(k \in \mathbb{Z})\\ \Rightarrow \text{TXĐ: } D=\mathbb{R} \setminus \left\{ \dfrac{\pi}{2}+k \pi;k \in \mathbb{Z}\right\}\\ 2)\\ a)\sin x =\dfrac{1}{2}\\ \sin x =\sin \dfrac{\pi}{6}\\ \left[\begin{array}{l} x =\dfrac{\pi}{6}+k 2 \pi (k \in \mathbb{Z}) \\ x =\dfrac{5\pi}{6}+k 2 \pi (k \in \mathbb{Z})\end{array} \right.\\ b)2\cos x+\sqrt{2}=0\\ \Leftrightarrow 2\cos x=-\sqrt{2}\\ \Leftrightarrow \cos x=-\dfrac{\sqrt{2}}{2}\\ \Leftrightarrow \cos x=\cos \dfrac{3\pi}{4}\\ \Leftrightarrow x=\pm \dfrac{3\pi}{4}+k 2 \pi (k \in \mathbb{Z})\\ c)\cos^2x+5\cos x+4=0\\ \Leftrightarrow \cos^2x+\cos x+4\cos x+4=0\\ \Leftrightarrow \cos x(\cos x+1)+4(\cos x+1)=0\\ \Leftrightarrow (\cos x+4)(\cos x+1)=0\\ \Leftrightarrow \left[\begin{array}{l} \cos x =-4(\text{Vô $$ nghiệm})\\ \cos x=-1\end{array} \right.\\ \Leftrightarrow \cos x=-1\\ \Leftrightarrow x=\pi+k 2 \pi (k \in \mathbb{Z})\\ d)\sin^2x-\cos x+1=0\\ \Leftrightarrow 1-\cos^2x-\cos x+1=0\\ \Leftrightarrow -\cos^2x-\cos x+2=0\\ \Leftrightarrow \cos^2x+\cos x-2=0\\ \Leftrightarrow \cos^2x-\cos x+2\cos x-2=0\\ \Leftrightarrow \cos x (\cos x-1)+2(\cos x-1)=0\\ \Leftrightarrow (\cos x+2) (\cos x-1)=0\\ \Leftrightarrow \left[\begin{array}{l} \cos x=-2(\text{Vô $$ nghiệm})\\ \cos x=1\end{array} \right.\\ \Leftrightarrow \cos x=1\\ \Leftrightarrow x=k 2 \pi (k \in \mathbb{Z}).$
