Đáp án:
1) \(\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)y = 0\\
\to \dfrac{{{{\cos }^2}x + \sin x.\cos x}}{{{{\sin }^2}x + 1}} = 0\\
\to {\cos ^2}x + \sin x.\cos x = 0\left( {do:{{\sin }^2}x + 1 > 0\forall x} \right)\\
\to \cos x\left( {\cos x + \sin x} \right) = 0\\
\to \left[ \begin{array}{l}
\cos x = 0\\
\cos x + \sin x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sin \left( {x + \dfrac{\pi }{4}} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x + \dfrac{\pi }{4} = k\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
2)DK:\sin x + \cos x \ne 2\\
\to \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) \ne 2\\
\to \sin \left( {x + \dfrac{\pi }{4}} \right) \ne \sqrt 2 \left( {ld} \right)\\
\to TXD:D = R\\
y = 0\\
\to \dfrac{{\cos x + 2}}{{\sin x + \cos x - 2}} = 0\\
\to \cos x + 2 = 0\left( {vô lý} \right)\\
Do:\cos x + 2 > 0\forall x\\
KL:x \in \emptyset
\end{array}\)
