Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
7,\\
\tan \left( {3x + \dfrac{\pi }{2}} \right).\cot \left( {5x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan \left( {3x + \dfrac{\pi }{2}} \right) = 0\\
\cot \left( {5x + 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{\pi }{2} = k\pi \\
5x + 1 = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\\
x = \dfrac{\pi }{{10}} - \dfrac{1}{5} + \dfrac{{k\pi }}{5}
\end{array} \right.\\
9,\\
5{\cos ^2}x + {\sin ^2}x = 4\\
\Leftrightarrow 5{\cos ^2}x + \left( {1 - {{\cos }^2}x} \right) = 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{{\sin }^2}x + {{\cos }^2}x = 1} \right)\\
\Leftrightarrow 4{\cos ^2}x = 3\\
\Leftrightarrow {\cos ^2}x = \dfrac{3}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{\sqrt 3 }}{2}\\
\cos x = - \dfrac{{\sqrt 3 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{\pi }{6} + k2\pi \\
x = \pm \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.
\end{array}\)
