1,tim nghiem cua x thuoc [0; PI/2] 2SIN ^{2}X-3SINX +1^{2}=0
2 câu trả lời
Đáp án:
\(pt\,\,\,co\,\,2\,\,\,nghiem\,\,\,x \in \left[ {0;\,\,\frac{\pi }{2}} \right]\,\,\,la:\,\,\,\,x \in \left\{ {\frac{\pi }{6};\,\,\frac{\pi }{2}} \right\}\) Hướng dẫn giải chi tiết: \[\begin{array}{l} 2{\sin ^2}x - 3\sin x + 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = 1\\ \sin x = \frac{1}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = \frac{\pi }{6} + m2\pi \\ x = \frac{{5\pi }}{6} + l2\pi \end{array} \right.\,\,\,\left( {k,\,\,m,\,\,l \in Z} \right)\\ 0 \le x \le \frac{\pi }{2} \Rightarrow \left[ \begin{array}{l} 0 \le \frac{\pi }{2} + k2\pi \le \frac{\pi }{2}\\ 0 \le \frac{\pi }{6} + m2\pi \le \frac{\pi }{2}\\ 0 \le \frac{{5\pi }}{6} + l2\pi \le \frac{\pi }{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - \frac{\pi }{2} \le k2\pi \le 0\\ - \frac{\pi }{6} \le m2\pi \le \frac{\pi }{3}\\ - \frac{{5\pi }}{6} \le l2\pi \le - \frac{\pi }{3} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - \frac{1}{4} \le k \le 0\\ - \frac{1}{{12}} \le m \le \frac{1}{6}\\ - \frac{5}{{12}} \le l \le - \frac{1}{6} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} k = 0\\ m = 0\\ l \in \emptyset \end{array} \right.\\ \Rightarrow pt\,\,\,co\,\,2\,\,\,nghiem\,\,\,x \in \left[ {0;\,\,\frac{\pi }{2}} \right]\,\,\,la:\,\,\,\,x \in \left\{ {\frac{\pi }{6};\,\,\frac{\pi }{2}} \right\}. \end{array}\]
Đáp án:
Giải thích các bước giải:
`2sin^2 x-3sin x+1^2=0`
`⇔` \(\left[ \begin{array}{l}sin x=1\\sin x=\dfrac{1}{2}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\\\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\end{array} \right.\)