Latifah 903 Questions 2k Answers 1 Best Answer 18 Points View Profile0 Latifah Asked: Tháng Mười Một 26, 20202020-11-26T17:15:29+00:00 2020-11-26T17:15:29+00:00In: Môn ToánGiúp mk mấy câu này vs!0Giúp mk mấy câu này vs! 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 AnswersOldestVotedRecentAcacia 930 Questions 2k Answers 0 Best Answers 16 Points View Profile Acacia 2020-11-26T17:17:13+00:00Added an answer on Tháng Mười Một 26, 2020 at 5:17 chiều Mình sử dụng công thức phương trình bậc hai nhé !!!`a) 3sin² 2x + 7cos 2x – 3 = 0``<=> 3 – 3cos² 2x + 7cos 2x – 3 = 0``<=>` \(\left[ \begin{array}{l}cos x = \dfrac{7}{3}\\cos x = 0\end{array} \right.\) `<=> x = π/2 + kπ (k ∈ ZZ)``b) 6cos² x + 5sin x – 7 = 0``<=> 6 – 6sin² x + 5sin x – 7 = 0``<=>` \(\left[ \begin{array}{l}sin x = \dfrac{1}{2}\\sin x = \dfrac{1}{3}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{6} + k2π\\x = \dfrac{5π}{6} + k2π\\x = arcsin \dfrac{1}{3} + k2π\\ x = π – arcsin \dfrac{1}{3} + k2π\end{array} \right.\) `(k ∈ ZZ)``c) cos 2x + cos x + 1 = 0``<=> 2cos² x – 1 + cos x + 1 = 0``<=>` \(\left[ \begin{array}{l}cos x = 0\\cos x = -\dfrac{1}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{2} + kπ\\x = ±\dfrac{2π}{3} + k2π\end{array} \right.\) `(k ∈ ZZ)``d) cos 2x – 5sin x – 3 = 0``<=> 1 – 2sin² x – 5sin x – 3 = 0``<=>` \(\left[ \begin{array}{l}sin x = -\dfrac{1}{2}\\sin x = -2 (l)\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = -\dfrac{π}{6} + k2π\\x = \dfrac{7π}{6} + k2π\end{array} \right.\) `(k ∈ ZZ)``e) 4sin^{4} x + 12cos² x = 7``<=> 4sin^{4} x + 12 – 12sin² x – 7 = 0``<=>` \(\left[ \begin{array}{l}sin² x = \dfrac{5}{2} (l)\\sin² x = \dfrac{1}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}sin x = \dfrac{1}{2}\\sin x = \dfrac{-1}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{6} + k2π\\x = \dfrac{5π}{6} + k2π\\x = \dfrac{-π}{6} + k2π\\x = \dfrac{7π}{6} + k2π\end{array} \right.\) `(k ∈ ZZ)`0Reply Share ShareShare on FacebookPhilomena 945 Questions 2k Answers 0 Best Answers 0 Points View Profile Philomena 2020-11-26T17:17:14+00:00Added an answer on Tháng Mười Một 26, 2020 at 5:17 chiều Giải thích các bước giải:Ta có:\(\begin{array}{l}a,\\3{\sin ^2}2x + 7\cos 2x – 3 = 0\\ \Leftrightarrow 3\left( {1 – {{\cos }^2}2x} \right) + 7\cos 2x – 3 = 0\\ \Leftrightarrow – 3{\cos ^2}2x + 7\cos 2x = 0\\ \Leftrightarrow \cos 2x\left( { – 3\cos 2x + 7} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = 0\\\cos 2x = \dfrac{7}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {L,\,\,\, – 1 \le \cos 2x \le 1} \right)\end{array} \right.\\ \Rightarrow \cos 2x = 0\\ \Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi \\ \Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\b,\\6{\cos ^2}x + 5\sin x – 7 = 0\\ \Leftrightarrow 6.\left( {1 – {{\sin }^2}x} \right) + 5\sin x – 7 = 0\\ \Leftrightarrow – 6{\sin ^2}x + 5\sin x – 1 = 0\\ \Leftrightarrow 6{\sin ^2}x – 5\sin x + 1 = 0\\ \Leftrightarrow \left( {2\sin x – 1} \right)\left( {3\sin x – 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = \dfrac{1}{2}\\\sin x = \dfrac{1}{3}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \\x = \arcsin \dfrac{1}{3} + k2\pi \\x = \pi – \arcsin \dfrac{1}{3} + k2\pi \end{array} \right.\\c,\\\cos 2x + \cos x + 1 = 0\\ \Leftrightarrow \left( {2{{\cos }^2}x – 1} \right) + \cos x + 1 = 0\\ \Leftrightarrow 2{\cos ^2}x + \cos x = 0\\ \Leftrightarrow \cos x\left( {2\cos x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\\\cos x = – \dfrac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\\d,\\\cos 2x – 5\sin x – 3 = 0\\ \Leftrightarrow \left( {1 – 2{{\sin }^2}x} \right) – 5\sin x – 3 = 0\\ \Leftrightarrow – 2{\sin ^2}x – 5\sin x – 2 = 0\\ \Leftrightarrow 2{\sin ^2}x + 5\sin x + 2 = 0\\ \Leftrightarrow \left( {\sin x + 2} \right)\left( {2\sin x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = – 2\\\sin x = – \dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \sin x = – \dfrac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { – 1 \le \sin x \le 1} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = – \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{7\pi }}{6} + k2\pi \end{array} \right.\\e,\\4{\sin ^4}x + 12{\cos ^2}x = 7\\ \Leftrightarrow 4.si{n^4}x + 12.\left( {1 – {{\sin }^2}x} \right) = 7\\ \Leftrightarrow 4{\sin ^4}x – 12{\sin ^2}x + 5 = 0\\ \Leftrightarrow \left( {2{{\sin }^2}x – 1} \right)\left( {2{{\sin }^2}x – 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}{\sin ^2}x = \dfrac{1}{2}\\{\sin ^2}x = \dfrac{5}{2}\end{array} \right.\\ – 1 \le \sin x \le 1 \Rightarrow 0 \le {\sin ^2}x \le 1\\ \Rightarrow {\sin ^2}x = \dfrac{1}{2}\\ \Leftrightarrow \sin x = \pm \dfrac{{\sqrt 2 }}{2}\\ \Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\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.
Acacia
Mình sử dụng công thức phương trình bậc hai nhé !!!
`a) 3sin² 2x + 7cos 2x – 3 = 0`
`<=> 3 – 3cos² 2x + 7cos 2x – 3 = 0`
`<=>` \(\left[ \begin{array}{l}cos x = \dfrac{7}{3}\\cos x = 0\end{array} \right.\)
`<=> x = π/2 + kπ (k ∈ ZZ)`
`b) 6cos² x + 5sin x – 7 = 0`
`<=> 6 – 6sin² x + 5sin x – 7 = 0`
`<=>` \(\left[ \begin{array}{l}sin x = \dfrac{1}{2}\\sin x = \dfrac{1}{3}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{6} + k2π\\x = \dfrac{5π}{6} + k2π\\x = arcsin \dfrac{1}{3} + k2π\\ x = π – arcsin \dfrac{1}{3} + k2π\end{array} \right.\) `(k ∈ ZZ)`
`c) cos 2x + cos x + 1 = 0`
`<=> 2cos² x – 1 + cos x + 1 = 0`
`<=>` \(\left[ \begin{array}{l}cos x = 0\\cos x = -\dfrac{1}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{2} + kπ\\x = ±\dfrac{2π}{3} + k2π\end{array} \right.\) `(k ∈ ZZ)`
`d) cos 2x – 5sin x – 3 = 0`
`<=> 1 – 2sin² x – 5sin x – 3 = 0`
`<=>` \(\left[ \begin{array}{l}sin x = -\dfrac{1}{2}\\sin x = -2 (l)\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = -\dfrac{π}{6} + k2π\\x = \dfrac{7π}{6} + k2π\end{array} \right.\) `(k ∈ ZZ)`
`e) 4sin^{4} x + 12cos² x = 7`
`<=> 4sin^{4} x + 12 – 12sin² x – 7 = 0`
`<=>` \(\left[ \begin{array}{l}sin² x = \dfrac{5}{2} (l)\\sin² x = \dfrac{1}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}sin x = \dfrac{1}{2}\\sin x = \dfrac{-1}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{6} + k2π\\x = \dfrac{5π}{6} + k2π\\x = \dfrac{-π}{6} + k2π\\x = \dfrac{7π}{6} + k2π\end{array} \right.\) `(k ∈ ZZ)`
Philomena
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
3{\sin ^2}2x + 7\cos 2x – 3 = 0\\
\Leftrightarrow 3\left( {1 – {{\cos }^2}2x} \right) + 7\cos 2x – 3 = 0\\
\Leftrightarrow – 3{\cos ^2}2x + 7\cos 2x = 0\\
\Leftrightarrow \cos 2x\left( { – 3\cos 2x + 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\cos 2x = \dfrac{7}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {L,\,\,\, – 1 \le \cos 2x \le 1} \right)
\end{array} \right.\\
\Rightarrow \cos 2x = 0\\
\Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
b,\\
6{\cos ^2}x + 5\sin x – 7 = 0\\
\Leftrightarrow 6.\left( {1 – {{\sin }^2}x} \right) + 5\sin x – 7 = 0\\
\Leftrightarrow – 6{\sin ^2}x + 5\sin x – 1 = 0\\
\Leftrightarrow 6{\sin ^2}x – 5\sin x + 1 = 0\\
\Leftrightarrow \left( {2\sin x – 1} \right)\left( {3\sin x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = \dfrac{1}{2}\\
\sin x = \dfrac{1}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = \arcsin \dfrac{1}{3} + k2\pi \\
x = \pi – \arcsin \dfrac{1}{3} + k2\pi
\end{array} \right.\\
c,\\
\cos 2x + \cos x + 1 = 0\\
\Leftrightarrow \left( {2{{\cos }^2}x – 1} \right) + \cos x + 1 = 0\\
\Leftrightarrow 2{\cos ^2}x + \cos x = 0\\
\Leftrightarrow \cos x\left( {2\cos x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x = – \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
d,\\
\cos 2x – 5\sin x – 3 = 0\\
\Leftrightarrow \left( {1 – 2{{\sin }^2}x} \right) – 5\sin x – 3 = 0\\
\Leftrightarrow – 2{\sin ^2}x – 5\sin x – 2 = 0\\
\Leftrightarrow 2{\sin ^2}x + 5\sin x + 2 = 0\\
\Leftrightarrow \left( {\sin x + 2} \right)\left( {2\sin x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = – 2\\
\sin x = – \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \sin x = – \dfrac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { – 1 \le \sin x \le 1} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
e,\\
4{\sin ^4}x + 12{\cos ^2}x = 7\\
\Leftrightarrow 4.si{n^4}x + 12.\left( {1 – {{\sin }^2}x} \right) = 7\\
\Leftrightarrow 4{\sin ^4}x – 12{\sin ^2}x + 5 = 0\\
\Leftrightarrow \left( {2{{\sin }^2}x – 1} \right)\left( {2{{\sin }^2}x – 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x = \dfrac{1}{2}\\
{\sin ^2}x = \dfrac{5}{2}
\end{array} \right.\\
– 1 \le \sin x \le 1 \Rightarrow 0 \le {\sin ^2}x \le 1\\
\Rightarrow {\sin ^2}x = \dfrac{1}{2}\\
\Leftrightarrow \sin x = \pm \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}
\end{array}\)