Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f,\\
3\sin x + \cos 2x = 2\\
\Leftrightarrow 3\sin x + \left( {1 - 2{{\sin }^2}x} \right) = 2\\
\Leftrightarrow 3\sin x + 1 - 2{\sin ^2}x - 2 = 0\\
\Leftrightarrow - 2{\sin ^2}x + 3\sin x - 1 = 0\\
\Leftrightarrow 2{\sin ^2}x - 3\sin x + 1 = 0\\
\Leftrightarrow \left( {\sin x - 1} \right)\left( {2\sin x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
g,\\
{\left( {\sin x + \cos x} \right)^2} = 1 + \cos x\\
\Leftrightarrow {\sin ^2}x + 2\sin x.\cos x + {\cos ^2}x = 1 + \cos x\\
\Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 2\sin x.\cos x = 1 + \cos x\\
\Leftrightarrow 1 + 2\sin x.\cos x = 1 + \cos x\\
\Leftrightarrow 2\sin x.\cos x - \cos x = 0\\
\Leftrightarrow \cos x.\left( {2\sin x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
2 câu bên dưới bị mất đề rồi em ạ
