Đáp án:
\(\begin{array}{l}
a)\,\,\left[ \begin{array}{l}
x = - \frac{\pi }{{12}} + k\pi \\
x = \frac{{7\pi }}{{12}} + k\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right).\\
b)\,\,x = k2\pi \,\,\,\left( {k \in Z} \right).\\
c)\,\,x = k\pi \,\,\,\,\left( {tm} \right).
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\,\,2\sin 2x + 1 = 0 \Leftrightarrow 2\sin 2x = - 1\\
\Leftrightarrow \sin 2x = - \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}
2x = - \frac{\pi }{6} + k2\pi \\
2x = \frac{{7\pi }}{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \frac{\pi }{{12}} + k\pi \\
x = \frac{{7\pi }}{{12}} + k\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right).\\
b)\,\,5{\cos ^2}x - 12\cos x + 7 = 0\\
\Leftrightarrow \left( {\cos x - 1} \right)\left( {5\cos x - 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x - 1 = 0\\
5\cos x - 7 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = \frac{7}{5}\,\,\,\left( {ktm} \right)
\end{array} \right. \Leftrightarrow x = k2\pi \,\,\,\left( {k \in Z} \right).\\
c)\,\,\,\tan \,\left( {x + \frac{\pi }{4}} \right) - 1 = 0\\
DK:\,\,\,\cos \left( {x + \frac{\pi }{4}} \right) \ne 0\\
\Leftrightarrow x + \frac{\pi }{4} \ne \frac{\pi }{2} + k\pi \\
\Leftrightarrow x \ne \frac{\pi }{4} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
pt \Leftrightarrow \tan \,\left( {x + \frac{\pi }{4}} \right) = 1\\
\Leftrightarrow x + \frac{\pi }{4} = \frac{\pi }{4} + k\pi \\
\Leftrightarrow x = k\pi \,\,\,\,\left( {tm} \right)
\end{array}\)
