Đáp án:
\(\begin{array}{l}
5,\\
x = \dfrac{\pi }{2} + k2\pi \,\,\,\,\left( {k \in Z} \right)\\
7,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{3}\\
x = \dfrac{1}{6}\arcsin \dfrac{1}{3} + \dfrac{{k\pi }}{3}\\
x = \dfrac{\pi }{6} - \dfrac{1}{6}\arcsin \dfrac{1}{3} + \dfrac{{k\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
5,\\
3{\cos ^2}x - 2\sin x + 2 = 0\\
\Leftrightarrow 3.\left( {1 - {{\sin }^2}x} \right) - 2\sin x + 2 = 0\\
\Leftrightarrow 3 - 3{\sin ^2}x - 2\sin x + 5 = 0\\
\Leftrightarrow - 3{\sin ^2}x - 2\sin x + 5 = 0\\
\Leftrightarrow 3{\sin ^2}x + 2\sin x - 5 = 0\\
\Leftrightarrow \left( {\sin x - 1} \right)\left( {3\sin x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - 1 = 0\\
3\sin x + 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = - \dfrac{5}{3}
\end{array} \right.\\
- 1 \le \sin x \le 1 \Rightarrow \sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \,\,\,\,\left( {k \in Z} \right)\\
7,\\
3{\cos ^2}6x + 4\sin 6x - 4 = 0\\
\Leftrightarrow 3.\left( {1 - {{\sin }^2}6x} \right) + 4\sin 6x - 4 = 0\\
\Leftrightarrow 3 - 3{\sin ^2}6x + 4\sin 6x - 4 = 0\\
\Leftrightarrow - 3{\sin ^2}6x + 4\sin 6x - 1 = 0\\
\Leftrightarrow 3{\sin ^2}6x - 4\sin 6x + 1 = 0\\
\Leftrightarrow \left( {\sin 6x - 1} \right)\left( {3\sin 6x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 6x - 1 = 0\\
3\sin 6x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin 6x = 1\\
\sin 6x = \dfrac{1}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
6x = \dfrac{\pi }{2} + k2\pi \\
6x = \arcsin \dfrac{1}{3} + k2\pi \\
6x = \pi - \arcsin \dfrac{1}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{3}\\
x = \dfrac{1}{6}\arcsin \dfrac{1}{3} + \dfrac{{k\pi }}{3}\\
x = \dfrac{\pi }{6} - \dfrac{1}{6}\arcsin \dfrac{1}{3} + \dfrac{{k\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
