Đáp án:
\[D = R\]
Giải thích các bước giải:
\(\begin{array}{l}
DKXD:\,\,\,{\sin ^2}x - 2\cos x + 4 \ne 0\\
\Leftrightarrow \left( {1 - {{\cos }^2}x} \right) - 2\cos x + 4 \ne 0\\
\Leftrightarrow 1 - {\cos ^2}x - 2\cos x + 4 \ne 0\\
\Leftrightarrow - {\cos ^2}x - 2\cos x + 5 \ne 0\\
\Leftrightarrow {\cos ^2}x + 2\cos x - 5 \ne 0\\
\Leftrightarrow \left( {{{\cos }^2}x + 2\cos x + 1} \right) - 6 \ne 0\\
\Leftrightarrow {\left( {\cos x + 1} \right)^2} \ne 6\\
\Leftrightarrow \left\{ \begin{array}{l}
\cos x + 1 \ne \sqrt 6 \\
\cos x + 1 \ne - \sqrt 6
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\cos x \ne \sqrt 6 - 1\\
\cos x \ne - \sqrt 6 - 1
\end{array} \right.,\,\,\,\forall x\,\,\,\left( { - 1 \le \cos x \le 1} \right)\\
Vậy\,\,\,TXD:\,\,\,D = R
\end{array}\)
