Đáp án:
\(x = - a + k\pi \left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
\text{Điều kiện: }\left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne k\pi \\
x \ne \dfrac{\pi }{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
Pt \to 2\left( {\tan x - \sin x + 1} \right) + 3\left( {\cot x - \cos x + 1} \right) = 0\\
\to 2\left( {\dfrac{{\sin x}}{{\cos x}} - \sin x + 1} \right) + 3\left( {\dfrac{{\cos x}}{{\sin x}} - \cos x + 1} \right) = 0\\
\to \dfrac{{2\sin x - 2\sin x\cos x + 2\cos x}}{{\cos x}} + \dfrac{{3\cos x - 3\sin x\cos x + 3\sin x}}{{\sin x}} = 0\\
\to \dfrac{{2\sin x\left( {\sin x - \sin x\cos x + \cos x} \right) + 3\cos x\left( {\sin x - \sin x\cos x + \cos x} \right)}}{{\sin x.\cos x}} = 0\\
\to 2\sin x\left( {\sin x - \sin x\cos x + \cos x} \right) + 3\cos x\left( {\sin x - \sin x\cos x + \cos x} \right) = 0\\
\to \left[ \begin{array}{l}
\sin x - \sin x\cos x + \cos x = 0\left( {\text{vô nghiệm}} \right)\\
2\sin x + 3\cos x = 0
\end{array} \right.\\
\to 2\sin x + 3\cos x = 0\\
\to \dfrac{2}{{\sqrt {13} }}\sin x + \dfrac{3}{{\sqrt {13} }}\cos x = 0\\
\text{Đặt: }\left\{ \begin{array}{l}
\dfrac{2}{{\sqrt {13} }} = \cos a\\
\dfrac{3}{{\sqrt {13} }} = \sin a
\end{array} \right.\\
\to \cos a.\sin x + \sin a.\cos x = 0\\
\to \sin \left( {x + a} \right) = 0\\
\to x + a = k\pi \\
\to x = - a + k\pi \left( {k \in Z} \right)
\end{array}\)
