1,tim nghiem cua x thuoc [0; PI/2] 2SIN ^{2}X-3SINX +1^{2}=0 24/09/2021 Bởi Mary 1,tim nghiem cua x thuoc [0; PI/2] 2SIN ^{2}X-3SINX +1^{2}=0
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: \[\begin{array}{l} 2{\sin ^2}x – 3\sin x + 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = 1\\ \sin x = \frac{1}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = \frac{\pi }{6} + m2\pi \\ x = \frac{{5\pi }}{6} + l2\pi \end{array} \right.\,\,\,\left( {k,\,\,m,\,\,l \in Z} \right)\\ 0 \le x \le \frac{\pi }{2} \Rightarrow \left[ \begin{array}{l} 0 \le \frac{\pi }{2} + k2\pi \le \frac{\pi }{2}\\ 0 \le \frac{\pi }{6} + m2\pi \le \frac{\pi }{2}\\ 0 \le \frac{{5\pi }}{6} + l2\pi \le \frac{\pi }{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} – \frac{\pi }{2} \le k2\pi \le 0\\ – \frac{\pi }{6} \le m2\pi \le \frac{\pi }{3}\\ – \frac{{5\pi }}{6} \le l2\pi \le – \frac{\pi }{3} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} – \frac{1}{4} \le k \le 0\\ – \frac{1}{{12}} \le m \le \frac{1}{6}\\ – \frac{5}{{12}} \le l \le – \frac{1}{6} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} k = 0\\ m = 0\\ l \in \emptyset \end{array} \right.\\ \Rightarrow pt\,\,\,co\,\,2\,\,\,nghiem\,\,\,x \in \left[ {0;\,\,\frac{\pi }{2}} \right]\,\,\,la:\,\,\,\,x \in \left\{ {\frac{\pi }{6};\,\,\frac{\pi }{2}} \right\}. \end{array}\] Bình luận
Đáp án: Giải thích các bước giải: `2sin^2 x-3sin x+1^2=0` `⇔` \(\left[ \begin{array}{l}sin x=1\\sin x=\dfrac{1}{2}\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\\\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\end{array} \right.\) Bình luận
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`
`⇔` \(\left[ \begin{array}{l}sin x=1\\sin x=\dfrac{1}{2}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\\\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\end{array} \right.\)