giải các pt lượng giác sau: a/tan3x.cot5x=1 b/2sin(x+π/5) — √3=0 c/cot(x+ π/4)-1=0 10/08/2021 Bởi Valentina giải các pt lượng giác sau: a/tan3x.cot5x=1 b/2sin(x+π/5) — √3=0 c/cot(x+ π/4)-1=0
Đáp án: $\begin{array}{l}a)\tan 3x.\cot 5x = 1\\\left( {dkxd:\left\{ \begin{array}{l}\cos 3x \ne 0\\\sin 5x \ne 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x \ne \frac{\pi }{6} + \frac{{k\pi }}{3}\\x \ne \frac{{k\pi }}{5}\end{array} \right.} \right)\\pt \Rightarrow \tan 3x = \frac{1}{{\cot 5x}}\\ \Rightarrow \tan 3x = \tan 5x\\ \Rightarrow 3x = 5x + k\pi \\ \Rightarrow x = \frac{{ – k\pi }}{2}\left( {ktmdk:x \ne \frac{\pi }{6} + \frac{{k\pi }}{3}} \right)\\Vậy\,x \in \emptyset \\b)2\sin \left( {x + \frac{\pi }{5}} \right) – \sqrt 3 = 0\\ \Rightarrow \sin \left( {x + \frac{\pi }{5}} \right) = \frac{{\sqrt 3 }}{2}\\ \Rightarrow \sin \left( {x + \frac{\pi }{5}} \right) = \sin \frac{\pi }{3}\\ \Rightarrow \left[ \begin{array}{l}x + \frac{\pi }{5} = \frac{\pi }{3} + k2\pi \\x + \frac{\pi }{5} = \pi – \frac{\pi }{3} + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = \frac{{2\pi }}{{15}} + k2\pi \\x = \frac{{7\pi }}{{15}} + k2\pi \end{array} \right.\\c)\cot \left( {x + \frac{\pi }{4}} \right) = 1\\ \Rightarrow x + \frac{\pi }{4} = \frac{\pi }{4} + k\pi \\ \Rightarrow x = k\pi \end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)\tan 3x.\cot 5x = 1\\
\left( {dkxd:\left\{ \begin{array}{l}
\cos 3x \ne 0\\
\sin 5x \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne \frac{\pi }{6} + \frac{{k\pi }}{3}\\
x \ne \frac{{k\pi }}{5}
\end{array} \right.} \right)\\
pt \Rightarrow \tan 3x = \frac{1}{{\cot 5x}}\\
\Rightarrow \tan 3x = \tan 5x\\
\Rightarrow 3x = 5x + k\pi \\
\Rightarrow x = \frac{{ – k\pi }}{2}\left( {ktmdk:x \ne \frac{\pi }{6} + \frac{{k\pi }}{3}} \right)\\
Vậy\,x \in \emptyset \\
b)2\sin \left( {x + \frac{\pi }{5}} \right) – \sqrt 3 = 0\\
\Rightarrow \sin \left( {x + \frac{\pi }{5}} \right) = \frac{{\sqrt 3 }}{2}\\
\Rightarrow \sin \left( {x + \frac{\pi }{5}} \right) = \sin \frac{\pi }{3}\\
\Rightarrow \left[ \begin{array}{l}
x + \frac{\pi }{5} = \frac{\pi }{3} + k2\pi \\
x + \frac{\pi }{5} = \pi – \frac{\pi }{3} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{{2\pi }}{{15}} + k2\pi \\
x = \frac{{7\pi }}{{15}} + k2\pi
\end{array} \right.\\
c)\cot \left( {x + \frac{\pi }{4}} \right) = 1\\
\Rightarrow x + \frac{\pi }{4} = \frac{\pi }{4} + k\pi \\
\Rightarrow x = k\pi
\end{array}$
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