Đáp án:
\(\begin{array}{l}
1,\\
x = \dfrac{\pi }{2} + k2\pi \,\,\,\,\left( {k \in Z} \right)\\
2,\\
x = - \dfrac{\pi }{8} + k\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
{\sin ^2}x - 3\sin x + 2 = 0\\
\Leftrightarrow \left( {{{\sin }^2}x - \sin x} \right) + \left( { - 2\sin x + 2} \right) = 0\\
\Leftrightarrow \sin x\left( {\sin x - 1} \right) - 2\left( {\sin x - 1} \right) = 0\\
\Leftrightarrow \left( {\sin x - 1} \right)\left( {\sin x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - 1 = 0\\
\sin x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = 2
\end{array} \right.\\
- 1 \le \sin x \le 1 \Rightarrow \sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \,\,\,\,\left( {k \in Z} \right)\\
2,\\
\sin 2x - \cos 2x = - \sqrt 2 \\
\Leftrightarrow \dfrac{{\sqrt 2 }}{2}.\sin 2x - \dfrac{{\sqrt 2 }}{2}.\cos 2x = \dfrac{{\sqrt 2 }}{2}.\left( { - \sqrt 2 } \right)\\
\Leftrightarrow \sin 2x.\cos \dfrac{\pi }{4} - \cos 2x.\sin \dfrac{\pi }{4} = - 1\\
\Leftrightarrow \sin \left( {2x - \dfrac{\pi }{4}} \right) = - 1\\
\Leftrightarrow 2x - \dfrac{\pi }{4} = - \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow 2x = - \dfrac{\pi }{4} + k2\pi \\
\Leftrightarrow x = - \dfrac{\pi }{8} + k\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
