giải pt a) 2cos2x-cosx+2=0 b) 7tanx-4cotx=12 c) sin^2 2x+cos^2 3x=1 14/08/2021 Bởi Vivian giải pt a) 2cos2x-cosx+2=0 b) 7tanx-4cotx=12 c) sin^2 2x+cos^2 3x=1
Đáp án: \(\begin{array}{l}a)\,\,\,\left\{\begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\x = \arccos \dfrac{1}{4} + k2\pi \\x = – \arccos \dfrac{1}{4} + k2\pi \end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\b)\,\,\,\left\{\begin{array}{l}x = \arctan 2 + k\pi \,\,\\x = \arctan \left( { – \dfrac{2}{7}} \right) + k\pi \,\end{array} \right.\\c)\,\,\left\{\begin{array}{l}x = k2\pi \\x = \dfrac{{k2\pi }}{5}\\x = \dfrac{\pi }{5} + \dfrac{{k2\pi }}{5}\\x = – \pi + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\end{array}\) Giải thích các bước giải: $\begin{array}{l}a)\,\,2\cos 2x – \cos x + 2 = 0\\ \Leftrightarrow 2\left( {2{{\cos }^2}x – 1} \right) – \cos x + 2 = 0\\ \Leftrightarrow 4{\cos ^2}x – \cos x = 0\\ \Leftrightarrow \cos x\left( {4\cos x – 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\\\cos x = \dfrac{1}{4}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\x = \arccos \dfrac{1}{4} + k2\pi \\x = – \arccos \dfrac{1}{4} + k2\pi \end{array} \right.\,\,\,\,\left( {k \in Z} \right).\\b)\,\,7\tan x – 4\cot x = 12\\\text{Điều kiện: }\left\{ \begin{array}{l}\sin x \ne 0\\\cos x \ne 0\end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne \dfrac{{k\pi }}{2}\,\,\,\left( {k \in Z} \right).\\\text{Phương trình} \Leftrightarrow 7\tan x – \dfrac{4}{{\tan x}} – 12 = 0\\ \Leftrightarrow 7{\tan ^2}x – 12\tan x – 4 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\tan x = 2\\\tan x = – \dfrac{2}{7}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \arctan 2 + k\pi \,\,\,\left( {tm} \right)\\x = \arctan \left( { – \dfrac{2}{7}} \right) + k\pi \,\,\,\left( {tm} \right)\end{array} \right..\\c)\,\,{\sin ^2}2x + {\cos ^2}3x = 1\\ \Leftrightarrow {\cos ^2}3x = 1 – {\sin ^2}2x\\ \Leftrightarrow {\cos ^2}3x = {\cos ^2}2x\\ \Leftrightarrow \left[ \begin{array}{l}\cos 3x = \cos 2x\\\cos 3x = – \cos 2x\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\cos 3x = \cos 2x\\\cos 3x = \cos \left( {\pi – 2x} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}3x = 2x + k2\pi \\3x = – 2x + k2\pi \\3x = \pi – 2x + k2\pi \\3x = – \pi + 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \dfrac{{k2\pi }}{5}\\x = \dfrac{\pi }{5} + \dfrac{{k2\pi }}{5}\\x = – \pi + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right).\end{array}$ Bình luận
Đáp án:
\(\begin{array}{l}
a)\,\,\,\left\{\begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \arccos \dfrac{1}{4} + k2\pi \\
x = – \arccos \dfrac{1}{4} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b)\,\,\,\left\{\begin{array}{l}
x = \arctan 2 + k\pi \,\,\\
x = \arctan \left( { – \dfrac{2}{7}} \right) + k\pi \,
\end{array} \right.\\
c)\,\,\left\{\begin{array}{l}
x = k2\pi \\
x = \dfrac{{k2\pi }}{5}\\
x = \dfrac{\pi }{5} + \dfrac{{k2\pi }}{5}\\
x = – \pi + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
$\begin{array}{l}
a)\,\,2\cos 2x – \cos x + 2 = 0\\
\Leftrightarrow 2\left( {2{{\cos }^2}x – 1} \right) – \cos x + 2 = 0\\
\Leftrightarrow 4{\cos ^2}x – \cos x = 0\\
\Leftrightarrow \cos x\left( {4\cos x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x = \dfrac{1}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \arccos \dfrac{1}{4} + k2\pi \\
x = – \arccos \dfrac{1}{4} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right).\\
b)\,\,7\tan x – 4\cot x = 12\\
\text{Điều kiện: }\left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0
\end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne \dfrac{{k\pi }}{2}\,\,\,\left( {k \in Z} \right).\\
\text{Phương trình} \Leftrightarrow 7\tan x – \dfrac{4}{{\tan x}} – 12 = 0\\
\Leftrightarrow 7{\tan ^2}x – 12\tan x – 4 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = 2\\
\tan x = – \dfrac{2}{7}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \arctan 2 + k\pi \,\,\,\left( {tm} \right)\\
x = \arctan \left( { – \dfrac{2}{7}} \right) + k\pi \,\,\,\left( {tm} \right)
\end{array} \right..\\
c)\,\,{\sin ^2}2x + {\cos ^2}3x = 1\\
\Leftrightarrow {\cos ^2}3x = 1 – {\sin ^2}2x\\
\Leftrightarrow {\cos ^2}3x = {\cos ^2}2x\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 3x = \cos 2x\\
\cos 3x = – \cos 2x
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos 3x = \cos 2x\\
\cos 3x = \cos \left( {\pi – 2x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 2x + k2\pi \\
3x = – 2x + k2\pi \\
3x = \pi – 2x + k2\pi \\
3x = – \pi + 2x + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{{k2\pi }}{5}\\
x = \dfrac{\pi }{5} + \dfrac{{k2\pi }}{5}\\
x = – \pi + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right).
\end{array}$