Hưng Gia 422 Questions 546 Answers 0 Best Answers 15 Points View Profile0 Hưng Gia Asked: Tháng Mười 25, 20202020-10-25T00:53:54+00:00 2020-10-25T00:53:54+00:00In: Môn Toánsin4x – cos4x = √3. cos8x0sin4x – cos4x = √3. cos8x ShareFacebookRelated Questions Một hình thang có đáy lớn là 52cm ; đáy bé kém đáy lớn 16cm ; chiều cao kém đáy ... Useful news and important articles APROTININ FROM BOVINE LUNG CELL CULTURE купить онлайн1 AnswerOldestVotedRecentTryphena 858 Questions 2k Answers 0 Best Answers 19 Points View Profile Tryphena 2020-10-25T00:55:17+00:00Added an answer on Tháng Mười 25, 2020 at 12:55 sáng Đáp án:$x = \dfrac{\pi }{{16}} + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} + \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} – \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}\left( {k \in Z} \right)$Giải thích các bước giải:$\begin{array}{l}\sin 4x – \cos 4x = \sqrt 3 \cos 8x\\ \Leftrightarrow \sin 4x – \cos 4x – \sqrt 3 \cos 8x = 0\\ \Leftrightarrow \sin 4x – \cos 4x – \sqrt 3 \left( {{{\cos }^2}4x – {{\sin }^2}4x} \right) = 0\\ \Leftrightarrow \sin 4x – \cos 4x + \sqrt 3 \left( {\sin 4x – \cos 4x} \right)\left( {\sin 4x + \cos 4x} \right) = 0\\ \Leftrightarrow \left( {\sin 4x – \cos 4x} \right)\left( {1 + \sqrt 3 \left( {\sin 4x + \cos 4x} \right)} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin 4x – \cos 4x = 0\\\sin 4x + \cos 4x = \dfrac{{ – 1}}{{\sqrt 3 }}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\dfrac{1}{{\sqrt 2 }}\sin 4x – \dfrac{1}{{\sqrt 2 }}\cos 4x = 0\\\dfrac{1}{{\sqrt 2 }}\sin 4x + \dfrac{1}{{\sqrt 2 }}\cos 4x = \dfrac{{ – 1}}{{\sqrt 6 }}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos \left( {4x + \dfrac{\pi }{4}} \right) = 0\\\cos \left( {4x – \dfrac{\pi }{4}} \right) = \dfrac{{ – 1}}{{\sqrt 6 }}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}4x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\4x – \dfrac{\pi }{4} = \arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k2\pi \\4x – \dfrac{\pi }{4} = – \arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k2\pi \end{array} \right.\left( {k \in Z} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{{16}} + k\dfrac{\pi }{2}\\x = \dfrac{\pi }{{16}} + \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}\\x = \dfrac{\pi }{{16}} – \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}\end{array} \right.\left( {k \in Z} \right)\end{array}$Vậy phương trình có các họ nghiệm là:$x = \dfrac{\pi }{{16}} + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} + \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} – \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}\left( {k \in Z} \right)$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.
Tryphena
Đáp án:
$x = \dfrac{\pi }{{16}} + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} + \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} – \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}\left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
\sin 4x – \cos 4x = \sqrt 3 \cos 8x\\
\Leftrightarrow \sin 4x – \cos 4x – \sqrt 3 \cos 8x = 0\\
\Leftrightarrow \sin 4x – \cos 4x – \sqrt 3 \left( {{{\cos }^2}4x – {{\sin }^2}4x} \right) = 0\\
\Leftrightarrow \sin 4x – \cos 4x + \sqrt 3 \left( {\sin 4x – \cos 4x} \right)\left( {\sin 4x + \cos 4x} \right) = 0\\
\Leftrightarrow \left( {\sin 4x – \cos 4x} \right)\left( {1 + \sqrt 3 \left( {\sin 4x + \cos 4x} \right)} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 4x – \cos 4x = 0\\
\sin 4x + \cos 4x = \dfrac{{ – 1}}{{\sqrt 3 }}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{{\sqrt 2 }}\sin 4x – \dfrac{1}{{\sqrt 2 }}\cos 4x = 0\\
\dfrac{1}{{\sqrt 2 }}\sin 4x + \dfrac{1}{{\sqrt 2 }}\cos 4x = \dfrac{{ – 1}}{{\sqrt 6 }}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos \left( {4x + \dfrac{\pi }{4}} \right) = 0\\
\cos \left( {4x – \dfrac{\pi }{4}} \right) = \dfrac{{ – 1}}{{\sqrt 6 }}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\
4x – \dfrac{\pi }{4} = \arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k2\pi \\
4x – \dfrac{\pi }{4} = – \arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{16}} + k\dfrac{\pi }{2}\\
x = \dfrac{\pi }{{16}} + \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}\\
x = \dfrac{\pi }{{16}} – \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
Vậy phương trình có các họ nghiệm là:
$x = \dfrac{\pi }{{16}} + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} + \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} – \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}\left( {k \in Z} \right)$