Lời giải:Ta có: \(\sin x + \cos x - 1 = 2\sin x\cos x\)
\(\begin{array}{l}
\Leftrightarrow \sin x + \cos x = 1 + 2\sin x\cos x\\
\Leftrightarrow \sin x + \cos x = {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x\\
\Leftrightarrow \sin x + \cos x = {\left( {\sin x + \cos x} \right)^2}
\end{array}\)
\( \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {1 - \sin x - \cos x} \right) = 0\)
\(\begin{array}{l}
\Leftrightarrow \left[ \begin{array}{l}
\sin x + \cos x = 0\\
1 - \sin x - \cos x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = - \cos x\\
\sin x + \cos x = 1
\end{array} \right.
\end{array}\)
\( \Leftrightarrow \left[ \begin{array}{l}\tan x = - 1\\\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\end{array} \right. \) \(\Leftrightarrow \left[ \begin{array}{l}x = - \dfrac{\pi }{4} + k\pi \\x = k2\pi \\x = \dfrac{\pi }{2} + k2\pi \end{array} \right.\,\left( {k \in \mathbb{Z}} \right)\)
Các nghiệm trên \(\left[ {0;2\pi } \right]\) là \(\left\{ {\dfrac{{3\pi }}{4};0;2\pi ;\dfrac{\pi }{2}} \right\}\)
Chọn đáp án C.
