Amity 926 Questions 2k Answers 0 Best Answers 13 Points View Profile0 Amity Asked: Tháng Mười Một 27, 20202020-11-27T07:29:32+00:00 2020-11-27T07:29:32+00:00In: Môn ToánGiải các phương trình bài 20Giải các phương trình bài 2 ShareFacebookRelated Questions Где быстро занять денег? Một hình thang có đáy lớn là 52cm ; đáy bé kém đáy lớn 16cm ; chiều cao kém đáy ... Useful news and important articles2 AnswersOldestVotedRecentEdana Edana 930 Questions 2k Answers 0 Best Answers 16 Points View Profile Edana 2020-11-27T07:30:42+00:00Added an answer on Tháng Mười Một 27, 2020 at 7:30 sáng `a) sin x = 1``<=> x = π/2 + k2π``b) cos 2x = -1``<=> 2x = π + k2π``<=> x = π/2 + kπ``c) tan 3x = 0``<=> 3x = kπ``<=> x = k(π)/2``d) cot 5x = -1``<=> 5x = -π/4 + kπ``<=> x = -π/20 + k(π)/5``e) sin (3x – π/3) = (\sqrt{3})/2``<=>` \(\left[ \begin{array}{l}3x – \dfrac{π}{3} = \dfrac{π}{3} + k2π\\3x – \dfrac{π}{3} = \dfrac{2π}{3} + k2π\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = \dfrac{2π}{9} + k\dfrac{2π}{3}\\x = \dfrac{π}{3} + k\dfrac{2π}{3}\end{array} \right.\) `(k ∈ ZZ)``f) cos (2x – π/4) = (\sqrt{2})/2``<=>` \(\left[ \begin{array}{l}2x – \dfrac{π}{4} = \dfrac{π}{4} + k2π\\2x – \dfrac{π}{4} = -\dfrac{π}{4} + k2π\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{4} + kπ\\x = kπ\end{array} \right.\) `(k ∈ ZZ)``g) tan ((3x)/2 + π/6) = sqrt{3}``<=> (3x)/2 + π/6 = π/3 + kπ``<=> (3x)/2 = π/6 + kπ``<=> x = π/9 + k(2π)/3` `(k ∈ ZZ)`0Reply Share ShareShare on FacebookNho 913 Questions 2k Answers 0 Best Answers 17 Points View Profile Nho 2020-11-27T07:31:07+00:00Added an answer on Tháng Mười Một 27, 2020 at 7:31 sáng Đáp án:$\begin{array}{l}1)\sin x = 1\\ \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \\2)\\\cos 2x = – 1\\ \Leftrightarrow 2x = \pi + k2\pi \\ \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\3)\tan 3x = 0\\ \Leftrightarrow 3x = k\pi \\ \Leftrightarrow x = \dfrac{{k\pi }}{3}\\4)cot5x = – 1\\ \Leftrightarrow 5x = – \dfrac{\pi }{4} + k\pi \\ \Leftrightarrow x = – \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{5}\\5)\sin \left( {3x – \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l}3x – \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\3x – \dfrac{\pi }{3} = \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}\\x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\end{array} \right.\\6)\cos \left( {2x – \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l}2x – \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\2x – \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{4} + k\pi \\x = \dfrac{\pi }{2} + k\pi \end{array} \right.\\7)\tan \left( {\dfrac{{3x}}{2} + \dfrac{\pi }{6}} \right) = \sqrt 3 \\ \Leftrightarrow \dfrac{{3x}}{2} + \dfrac{\pi }{6} = \dfrac{\pi }{3} + k\pi \\ \Leftrightarrow x = \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3}\end{array}$0Reply Share ShareShare on FacebookLeave an answerLeave an answerHủy By answering, you agree to the Terms of Service and Privacy Policy .* Lưu tên của tôi, email, và trang web trong trình duyệt này cho lần bình luận kế tiếp của tôi.
Edana Edana
`a) sin x = 1`
`<=> x = π/2 + k2π`
`b) cos 2x = -1`
`<=> 2x = π + k2π`
`<=> x = π/2 + kπ`
`c) tan 3x = 0`
`<=> 3x = kπ`
`<=> x = k(π)/2`
`d) cot 5x = -1`
`<=> 5x = -π/4 + kπ`
`<=> x = -π/20 + k(π)/5`
`e) sin (3x – π/3) = (\sqrt{3})/2`
`<=>` \(\left[ \begin{array}{l}3x – \dfrac{π}{3} = \dfrac{π}{3} + k2π\\3x – \dfrac{π}{3} = \dfrac{2π}{3} + k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \dfrac{2π}{9} + k\dfrac{2π}{3}\\x = \dfrac{π}{3} + k\dfrac{2π}{3}\end{array} \right.\) `(k ∈ ZZ)`
`f) cos (2x – π/4) = (\sqrt{2})/2`
`<=>` \(\left[ \begin{array}{l}2x – \dfrac{π}{4} = \dfrac{π}{4} + k2π\\2x – \dfrac{π}{4} = -\dfrac{π}{4} + k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{4} + kπ\\x = kπ\end{array} \right.\) `(k ∈ ZZ)`
`g) tan ((3x)/2 + π/6) = sqrt{3}`
`<=> (3x)/2 + π/6 = π/3 + kπ`
`<=> (3x)/2 = π/6 + kπ`
`<=> x = π/9 + k(2π)/3` `(k ∈ ZZ)`
Nho
Đáp án:
$\begin{array}{l}
1)\sin x = 1\\
\Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \\
2)\\
\cos 2x = – 1\\
\Leftrightarrow 2x = \pi + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
3)\tan 3x = 0\\
\Leftrightarrow 3x = k\pi \\
\Leftrightarrow x = \dfrac{{k\pi }}{3}\\
4)cot5x = – 1\\
\Leftrightarrow 5x = – \dfrac{\pi }{4} + k\pi \\
\Leftrightarrow x = – \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{5}\\
5)\sin \left( {3x – \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
3x – \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\
3x – \dfrac{\pi }{3} = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}
\end{array} \right.\\
6)\cos \left( {2x – \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x – \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
2x – \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
7)\tan \left( {\dfrac{{3x}}{2} + \dfrac{\pi }{6}} \right) = \sqrt 3 \\
\Leftrightarrow \dfrac{{3x}}{2} + \dfrac{\pi }{6} = \dfrac{\pi }{3} + k\pi \\
\Leftrightarrow x = \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3}
\end{array}$