Đáp án:
TXD: $D = R\backslash \left\{ {\pi + k\pi ;k \in Z} \right\}$
1/ $x = \pi + k\pi $
2/ \(\left[ \begin{array}{l}
x = {\mathop{\rm arccot}\nolimits} 3 + k\pi \\
x = {\mathop{\rm arccot}\nolimits} \dfrac{1}{2} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
y = \tan x + \cot x - 2\sin x\\
DK:\,\left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0
\end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow x \ne \pi + k\pi \\
D = R\backslash \left\{ {\pi + k\pi ;k \in Z} \right\}\\
1)\,\sin 2x + 1 = 6\sin x + 1 - 2{\sin ^2}x\\
\Leftrightarrow 2{\sin ^2}x - 6\sin x + 2\sin x\cos x = 0\\
\Leftrightarrow \sin x\left( {\sin x + \cos x - 3} \right) = 0\\
\Rightarrow \sin x = 0 \Leftrightarrow x = \pi + k\pi \\
2)\,\dfrac{3}{{\cot x + 1}} + \dfrac{1}{{2\cot x - 2}} = 0\,\,\left( {\cot x \ne \pm 1} \right)\\
\Rightarrow 6\cot x - 6 + \cot x + 1 = 2\left( {{{\cot }^2}x - 1} \right)\\
\Leftrightarrow 2{\cot ^2}x - 7\cot x + 3 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cot x = 3\\
\cot x = \dfrac{1}{2}
\end{array} \right.\left( {tm} \right) \Leftrightarrow \left[ \begin{array}{l}
x = {\mathop{\rm arccot}\nolimits} 3 + k\pi \\
x = {\mathop{\rm arccot}\nolimits} \dfrac{1}{2} + k\pi
\end{array} \right.
\end{array}\)
