Giải pt: a. $cos^22x=\frac{1}{4}$ b. $cos2x.tanx=0$ c. $ sin3x.cotx=0$ 26/07/2021 Bởi Daisy Giải pt: a. $cos^22x=\frac{1}{4}$ b. $cos2x.tanx=0$ c. $ sin3x.cotx=0$
Đáp án: a. \(\left[ \begin{array}{l}x = \dfrac{\pi }{6} + k\pi \\x = – \dfrac{\pi }{6} + k\pi \\x = \dfrac{\pi }{3} + k\pi \\x = – \dfrac{\pi }{3} + k\pi \end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.\left[ \begin{array}{l}\cos 2x = \dfrac{1}{2}\\\cos 2x = – \dfrac{1}{2}\end{array} \right.\\ \to \left[ \begin{array}{l}2x = \dfrac{\pi }{3} + k2\pi \\2x = – \dfrac{\pi }{3} + k2\pi \\2x = \dfrac{{2\pi }}{3} + k2\pi \\2x = – \dfrac{{2\pi }}{3} + k2\pi \end{array} \right. \to \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k\pi \\x = – \dfrac{\pi }{6} + k\pi \\x = \dfrac{\pi }{3} + k\pi \\x = – \dfrac{\pi }{3} + k\pi \end{array} \right.\left( {k \in Z} \right)\\b.DK:\cos x \ne 0 \to x \ne \dfrac{\pi }{2} + k\pi \\\left[ \begin{array}{l}\cos 2x = 0\\\tan x = 0\end{array} \right. \to \left[ \begin{array}{l}2x = \dfrac{\pi }{2} + k\pi \\x = k\pi \end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\x = k\pi \end{array} \right.\\ \to x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\c.DK:\sin x \ne 0 \to x \ne k\pi \\\sin 3x.\cot x = 0\\ \to \left[ \begin{array}{l}\sin 3x = 0\\\cos x = 0\end{array} \right.\\ \to \left[ \begin{array}{l}3x = k\pi \\x = \dfrac{\pi }{2} + k\pi \end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{{k\pi }}{3}\\x = \dfrac{\pi }{2} + k\pi \end{array} \right.\left( {k \in Z} \right)\end{array}\) Bình luận
Đáp án: `a)` \(\left[ \begin{array}{l}x = ±\frac{π}{6}\\x = ±\frac{π}{3}\end{array} \right.\) `(k ∈ ZZ)` `b)` \(\left[ \begin{array}{l}x = \frac{π}{4} + k\frac{π}{2}\\x = kπ\end{array} \right.\) `(k ∈ ZZ)` `c)` \(\left[ \begin{array}{l}x = k\frac{π}{3}\\x = \frac{π}{2} + kπ\end{array} \right.\) `(k ∈ ZZ)` Giải thích các bước giải: `a) cos² 2x = 1/4` `<=>` \(\left[ \begin{array}{l}cos 2x = frac{1}{2}\\cos 2x = -\frac{1}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}2x = ±\frac{π}{3}\\2x = ±\frac{2π}{3}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = ±\frac{π}{6}\\x = ±\frac{π}{3}\end{array} \right.\) `(k ∈ ZZ)` `b) cos 2x.tan x = 0` `ĐKXĐ: x ne (π)/2 + kπ` `=>` \(\left[ \begin{array}{l}cos 2x = 0\\tan x = 0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}2x = \frac{π}{2} + kπ\\sin x = 0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = \frac{π}{4} + k\frac{π}{2}\\x = kπ\end{array} \right.\) `(k ∈ ZZ)` `c) sin 3x.cot x = 0` `ĐKXĐ: x ne kπ` `=>` \(\left[ \begin{array}{l}sin 3x = 0\\cot x = 0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}3x = kπ\\x = \frac{π}{2} + kπ\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = k\frac{π}{3}\\x = \frac{π}{2} + kπ\end{array} \right.\) `(k ∈ ZZ)` Bình luận
Đáp án:
a. \(\left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x = – \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{3} + k\pi \\
x = – \dfrac{\pi }{3} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left[ \begin{array}{l}
\cos 2x = \dfrac{1}{2}\\
\cos 2x = – \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = \dfrac{\pi }{3} + k2\pi \\
2x = – \dfrac{\pi }{3} + k2\pi \\
2x = \dfrac{{2\pi }}{3} + k2\pi \\
2x = – \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x = – \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{3} + k\pi \\
x = – \dfrac{\pi }{3} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
b.DK:\cos x \ne 0 \to x \ne \dfrac{\pi }{2} + k\pi \\
\left[ \begin{array}{l}
\cos 2x = 0\\
\tan x = 0
\end{array} \right. \to \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
x = k\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = k\pi
\end{array} \right.\\
\to x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
c.DK:\sin x \ne 0 \to x \ne k\pi \\
\sin 3x.\cot x = 0\\
\to \left[ \begin{array}{l}
\sin 3x = 0\\
\cos x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
Đáp án:
`a)` \(\left[ \begin{array}{l}x = ±\frac{π}{6}\\x = ±\frac{π}{3}\end{array} \right.\) `(k ∈ ZZ)`
`b)` \(\left[ \begin{array}{l}x = \frac{π}{4} + k\frac{π}{2}\\x = kπ\end{array} \right.\) `(k ∈ ZZ)`
`c)` \(\left[ \begin{array}{l}x = k\frac{π}{3}\\x = \frac{π}{2} + kπ\end{array} \right.\) `(k ∈ ZZ)`
Giải thích các bước giải:
`a) cos² 2x = 1/4`
`<=>` \(\left[ \begin{array}{l}cos 2x = frac{1}{2}\\cos 2x = -\frac{1}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x = ±\frac{π}{3}\\2x = ±\frac{2π}{3}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = ±\frac{π}{6}\\x = ±\frac{π}{3}\end{array} \right.\) `(k ∈ ZZ)`
`b) cos 2x.tan x = 0`
`ĐKXĐ: x ne (π)/2 + kπ`
`=>` \(\left[ \begin{array}{l}cos 2x = 0\\tan x = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x = \frac{π}{2} + kπ\\sin x = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \frac{π}{4} + k\frac{π}{2}\\x = kπ\end{array} \right.\) `(k ∈ ZZ)`
`c) sin 3x.cot x = 0`
`ĐKXĐ: x ne kπ`
`=>` \(\left[ \begin{array}{l}sin 3x = 0\\cot x = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x = kπ\\x = \frac{π}{2} + kπ\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = k\frac{π}{3}\\x = \frac{π}{2} + kπ\end{array} \right.\) `(k ∈ ZZ)`