Bài 1: A/sin^2(2x)=1/2 B/cot^2(x/2)=1/3 C/tan(π/12+12x)=-căn3 11/07/2021 Bởi Rylee Bài 1: A/sin^2(2x)=1/2 B/cot^2(x/2)=1/3 C/tan(π/12+12x)=-căn3
Đáp án: a. \(\left[ \begin{array}{l}x = \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{{5\pi }}{{12}} + k\pi \\x = – \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{{7\pi }}{{12}} + k\pi \end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.{\sin ^2}\left( {2x} \right) = \dfrac{1}{2}\\ \to \left[ \begin{array}{l}\sin 2x = \dfrac{1}{2}\\\sin 2x = – \dfrac{1}{2}\end{array} \right.\\ \to \left[ \begin{array}{l}2x = \dfrac{\pi }{6} + k2\pi \\2x = \dfrac{{5\pi }}{6} + k2\pi \\2x = – \dfrac{\pi }{6} + k2\pi \\2x = \dfrac{{7\pi }}{6} + k2\pi \end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{{5\pi }}{{12}} + k\pi \\x = – \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{{7\pi }}{{12}} + k\pi \end{array} \right.\left( {k \in Z} \right)\\b.DK:\sin \left( {\dfrac{x}{2}} \right) \ne 0\\{\cot ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{3}\\ \to {\tan ^2}\dfrac{x}{2} = 3\\ \to \left[ \begin{array}{l}\tan \dfrac{x}{2} = \sqrt 3 \\\tan \dfrac{x}{2} = – \sqrt 3 \end{array} \right.\\ \to \left[ \begin{array}{l}\dfrac{x}{2} = \dfrac{\pi }{3} + k\pi \\\dfrac{x}{2} = – \dfrac{\pi }{3} + k\pi \end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{{2\pi }}{3} + k2\pi \\x = – \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)\\c.DK:\cos \left( {\dfrac{\pi }{{12}} + 12x} \right) \ne 0\\\tan \left( {\dfrac{\pi }{{12}} + 12x} \right) = – \sqrt 3 \\ \to \dfrac{\pi }{{12}} + 12x = – \dfrac{\pi }{3} + k\pi \\ \to 12x = – \dfrac{{5\pi }}{{12}} + k\pi \\ \to x = – 5\pi + k12\pi \left( {k \in Z} \right)\end{array}\) Bình luận
Đáp án:
a. \(\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{5\pi }}{{12}} + k\pi \\
x = – \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{7\pi }}{{12}} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.{\sin ^2}\left( {2x} \right) = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
\sin 2x = \dfrac{1}{2}\\
\sin 2x = – \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{{5\pi }}{6} + k2\pi \\
2x = – \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{5\pi }}{{12}} + k\pi \\
x = – \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{7\pi }}{{12}} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
b.DK:\sin \left( {\dfrac{x}{2}} \right) \ne 0\\
{\cot ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{3}\\
\to {\tan ^2}\dfrac{x}{2} = 3\\
\to \left[ \begin{array}{l}
\tan \dfrac{x}{2} = \sqrt 3 \\
\tan \dfrac{x}{2} = – \sqrt 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{x}{2} = \dfrac{\pi }{3} + k\pi \\
\dfrac{x}{2} = – \dfrac{\pi }{3} + k\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = – \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
c.DK:\cos \left( {\dfrac{\pi }{{12}} + 12x} \right) \ne 0\\
\tan \left( {\dfrac{\pi }{{12}} + 12x} \right) = – \sqrt 3 \\
\to \dfrac{\pi }{{12}} + 12x = – \dfrac{\pi }{3} + k\pi \\
\to 12x = – \dfrac{{5\pi }}{{12}} + k\pi \\
\to x = – 5\pi + k12\pi \left( {k \in Z} \right)
\end{array}\)