Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
d,\\
{\sin ^2}x + \cos 2x + \cos x = 2\\
\Leftrightarrow \left( {1 - {{\cos }^2}x} \right) + \left( {2{{\cos }^2}x - 1} \right) + \cos x = 2\\
\Leftrightarrow {\cos ^2}x + \cos x - 2 = 0\\
\Leftrightarrow \left( {\cos x - 1} \right)\left( {\cos x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = - 2
\end{array} \right.\\
\Leftrightarrow \cos x = 1\\
\Leftrightarrow x = k2\pi \\
e,\\
\cos 2x + {\cos ^2}x - \sin x + 2 = 0\\
\Leftrightarrow \left( {1 - 2{{\sin }^2}x} \right) + \left( {1 - {{\sin }^2}x} \right) - \sin x + 2 = 0\\
\Leftrightarrow - 3{\sin ^2}x - \sin x + 4 = 0\\
\Leftrightarrow 3{\sin ^2} + \sin x - 4 = 0\\
\Leftrightarrow \left( {\sin x - 1} \right)\left( {3\sin x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = - \dfrac{4}{3}\,\,\,\,\left( L \right)
\end{array} \right.\\
\Leftrightarrow \sin x = 1\\
\Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \\
f,\\
3{\cos ^2}x - 2\cos 2x = 3\sin x - 1\\
\Leftrightarrow 3.\left( {1 - {{\sin }^2}x} \right) - 2.\left( {1 - 2{{\sin }^2}x} \right) = 3\sin x - 1\\
\Leftrightarrow {\sin ^2}x + 1 = 3\sin x - 1\\
\Leftrightarrow {\sin ^2}x - 3\sin x + 2 = 0\\
\Leftrightarrow \left( {\sin x - 1} \right)\left( {\sin x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = 2
\end{array} \right.\\
\Leftrightarrow \sin x = 1\\
\Leftrightarrow x = \dfrac{\pi }{2} + k2\pi
\end{array}\)
