Giải thích các bước giải:
a,
Ta có:
\[\begin{array}{l}
\cos x - \cos 2x + \sin x = 0\\
\Leftrightarrow \left( {\cos x + \sin x} \right) - \left( {{{\cos }^2}x - {{\sin }^2}x} \right) = 0\\
\Leftrightarrow \left( {\cos x + \sin x} \right) - \left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) = 0\\
\Leftrightarrow \left( {\cos x + \sin x} \right)\left( {1 - \cos x + \sin x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x + \sin x = 0\\
\cos x - \sin x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = 0\\
\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = - 1
\end{array} \right.
\end{array}\]
b,
đk: \[\left\{ \begin{array}{l}
\tan x \ne - 1\\
\cos x \ne 0\\
\sin x \ne 0
\end{array} \right.\]
\[\begin{array}{l}
\frac{2}{{\tan x + 1}} + \frac{1}{{\tan x}} = 2\\
\Leftrightarrow \frac{2}{{\frac{{\sin x}}{{\cos x}} + 1}} + \frac{1}{{\frac{{\sin x}}{{\cos x}}}} = 2\\
\Leftrightarrow \frac{{2\cos x}}{{\sin x + \cos x}} + \frac{{\cos x}}{{\sin x}} = 2\\
\Leftrightarrow \frac{{2\cos x.\sin x + \cos x\left( {\sin x + \cos x} \right)}}{{\sin x\left( {\sin x + \cos x} \right)}} = 2\\
\Leftrightarrow 3\sin x\cos x + {\cos ^2}x = 2{\sin ^2}x + 2\sin x\cos x\\
\Leftrightarrow 2{\sin ^2}x - \sin x\cos x - {\cos ^2}x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = \cos x\\
\sin x = \frac{{ - 1}}{2}\cos x
\end{array} \right.
\end{array}\]
