Giải pt :
\(tan.\left(x+\dfrac{\pi}{4}\right)+cot\cdot\left(2x-\dfrac{\pi}{3}\right)=0\)
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AH
Akai Haruma
Giáo viên
28 tháng 6 2021
a1.
$\cot (2x+\frac{\pi}{3})=-\sqrt{3}=\cot \frac{-\pi}{6}$
$\Rightarrow 2x+\frac{\pi}{3}=\frac{-\pi}{6}+k\pi$ với $k$ nguyên
$\Leftrightarrow x=\frac{-\pi}{4}+\frac{k}{2}\pi$ với $k$ nguyên
a2. ĐKXĐ:...............
$\cot (3x-10^0)=\frac{1}{\cot 2x}=\tan 2x$
$\Leftrightarrow \cot (3x-\frac{\pi}{18})=\cot (\frac{\pi}{2}-2x)$
$\Rightarrow 3x-\frac{\pi}{18}=\frac{\pi}{2}-2x+k\pi$ với $k$ nguyên
$\Leftrightarrow x=\frac{\pi}{9}+\frac{k}{5}\pi$ với $k$ nguyên.
AH
Akai Haruma
Giáo viên
28 tháng 6 2021
a3. ĐKXĐ:........
$\cot (\frac{\pi}{4}-2x)-\tan x=0$
$\Leftrightarrow \cot (\frac{\pi}{4}-2x)=\tan x=\cot (\frac{\pi}{2}-x)$
$\Rightarrow \frac{\pi}{4}-2x=\frac{\pi}{2}-x+k\pi$ với $k$ nguyên
$\Leftrightarrow x=-\frac{\pi}{4}+k\pi$ với $k$ nguyên.
a4. ĐKXĐ:.....
$\cot (\frac{\pi}{6}+3x)+\tan (x-\frac{\pi}{18})=0$
$\Leftrightarrow \cot (\frac{\pi}{6}+3x)=-\tan (x-\frac{\pi}{18})=\tan (\frac{\pi}{18}-x)$
$=\cot (x+\frac{4\pi}{9})$
$\Rightarrow \frac{\pi}{6}+3x=x+\frac{4\pi}{9}+k\pi$ với $k$ nguyên
$\Rightarrow x=\frac{5}{36}\pi + \frac{k}{2}\pi$ với $k$ nguyên.
23 tháng 6 2021
a, Ta có : \(\sin\left(3x+60\right)=\dfrac{1}{2}\)
\(\Rightarrow3x+60=30+2k180\)
\(\Rightarrow3x=2k180-30\)
\(\Leftrightarrow x=120k-10\)
Vậy ...
b, Ta có : \(\cos\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow2x-\dfrac{\pi}{3}=\dfrac{3}{4}\pi+k2\pi\)
\(\Leftrightarrow x=\dfrac{13}{24}\pi+k\pi\)
Vậy ...
c, Ta có : \(tan\left(x+\dfrac{\pi}{6}\right)=\sqrt{3}\)
\(\Rightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)
Vậy ...
d, Ta có : \(\cot\left(2x+\pi\right)=-1\)
\(\Rightarrow2x+\pi=\dfrac{3}{4}\pi+k\pi\)
\(\Leftrightarrow x=-\dfrac{1}{8}\pi+\dfrac{k}{2}\pi\)
Vậy ...
23 tháng 6 2021
a) \(sin\left(3x+60^0\right)=\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{3}\right)=sin\dfrac{\pi}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(\(k\in Z\))
Vậy...
b) Pt\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\dfrac{3\pi}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13\pi}{24}+k\pi\\x=-\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)(\(k\in Z\))
Vậy...
c) Pt \(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=tan\dfrac{\pi}{3}\)
\(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi,k\in Z\)\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi,k\in Z\)
Vậy...
d) Pt \(\Leftrightarrow tan\left(2x+\pi\right)=-1\)
\(\Leftrightarrow2x+\pi=-\dfrac{\pi}{4}+k\pi,k\in Z\)
\(\Leftrightarrow x=-\dfrac{5\pi}{8}+\dfrac{k\pi}{2},k\in Z\)
Vậy...
18 tháng 5 2017
a) \(x=-45^0+k90^0,k\in\mathbb{Z}\)
b) \(x=-\dfrac{\pi}{6}+k\pi,k\in\mathbb{Z}\)
c) \(x=\dfrac{3\pi}{4}+k2\pi,k\in\mathbb{Z}\)
d) \(x=300^0+k540^0,k\in\mathbb{Z}\)
R
2
27 tháng 9 2021
Pt \(\Leftrightarrow\)\(tan\left(x+\dfrac{\pi}{3}\right)\)=\(-cot\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\)\(tan\left(x+\dfrac{\pi}{3}\right)\)=\(tan\left(\dfrac{\pi}{2}+\dfrac{\pi}{2}-3x\right)\)=\(tan\left(\pi-3x\right)\)
\(\Leftrightarrow\)\(x+\dfrac{\pi}{3}=\pi-3x+k\pi\)
\(\Leftrightarrow\)4\(x\)=\(\dfrac{4}{3}\pi+k\pi\)
\(\Leftrightarrow\) \(x=\) \(\dfrac{\pi}{3}+k\dfrac{\pi}{4}\)(\(k\in Z\))
HP
28 tháng 9 2021
\(pt\Leftrightarrow tan\left(x+\dfrac{\pi}{3}\right)=-cot\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow tan\left(x+\dfrac{\pi}{3}\right)=cot\left(-\dfrac{\pi}{2}+3x\right)\)
\(\Leftrightarrow tan\left(x+\dfrac{\pi}{3}\right)=tan\left(\dfrac{\pi}{2}+\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow tan\left(x+\dfrac{\pi}{3}\right)=tan\left(\pi-3x\right)\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=\pi-3x+k\pi\)
\(\Leftrightarrow4x=\dfrac{2\pi}{3}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{k\pi}{4}\)
HP
15 tháng 4 2021
\(P=\left[tan\dfrac{17\pi}{4}+tan\left(\dfrac{7\pi}{2}-x\right)\right]^2+\left[cot\dfrac{13\pi}{4}+cot\left(7\pi-x\right)\right]^2\)
\(=\left[tan\dfrac{\pi}{4}+tan\left(-\dfrac{\pi}{2}-x\right)\right]^2+\left[cot\left(-\dfrac{3\pi}{4}\right)+cot\left(-\pi-x\right)\right]^2\)
\(=\left[tan\dfrac{\pi}{4}-cotx\right]^2+\left[tan\dfrac{\pi}{4}-cotx\right]^2\)
\(=2\left(1-cotx\right)^2\)
AH
Akai Haruma
Giáo viên
20 tháng 12 2017
1)
Ta có:
\(\int (2-\cot ^2x)dx=\int (2-\frac{\cos ^2x}{\sin ^2x})dx\)
\(=\int (2-\frac{1-\sin ^2x}{\sin ^2x})dx=\int (3-\frac{1}{\sin ^2x})dx=3\int dx-\int \frac{dx}{\sin ^2x}\)
\(=3x+\int d(\cot x)=3x+\cot x+c\)
\(\Rightarrow \int ^{\frac{\pi}{2}}_{\frac{\pi}{3}}(2-\cot ^2x)dx=\left.\begin{matrix} \frac{\pi}{2}\\ \frac{\pi}{3}\end{matrix}\right|(3x+\cot x+c)=\frac{\pi}{2}-\frac{\sqrt{3}}{3}\)
3)
Xét \(\int (2\tan x-3\cot x)^2dx\)
\(=\int (4\tan ^2x+9\cot ^2x-12)dx\)
\(=\int (\frac{4\sin ^2x}{\cos ^2x}+\frac{9\cos ^2x}{\sin ^2x}-12)dx\)
\(=\int (\frac{4(1-\cos ^2x)}{\cos ^2x}+\frac{9(1-\sin ^2x)}{\sin ^2x}-12)dx\)
\(=\int (\frac{4}{\cos ^2x}+\frac{9}{\sin ^2x}-25)dx\)
\(=4\int d(\tan x)-9\int d(\cot x)-25\int dx\)
\(=4\tan x-9\cot x-25x+c\)
Do đó:
\(\int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}(2\tan x-3\cot x)^2dx=\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|(4\tan x-9\cot x-25x+c)=\frac{26\sqrt{3}}{3}-\frac{25\pi}{6}\)
AH
Akai Haruma
Giáo viên
20 tháng 12 2017
2)
Xét \(\int (\tan x+\cot x)^2dx=\int (\tan ^2x+\cot ^2x+2)dx\)
\(=\int (\frac{\sin ^2x}{\cos^2 x}+\frac{\cos ^2x}{\sin ^2x}+2)dx\)
\(=\int (\frac{1-\cos ^2x}{\cos ^2x}+\frac{1-\sin ^2x}{\sin ^2x}+2)dx\)
\(=\int (\frac{1}{\cos ^2x}+\frac{1}{\sin ^2x})dx\)
\(=\int d(\tan x)-\int d(\cot x)=\tan x-\cot x+c\)
Do đó:
\(\int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}(\tan x+\cot x)^2dx=\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|(\tan x-\cot x+c)=2\sqrt{3}-\frac{2\sqrt{3}}{3}\)
\(tan\cdot\left(x+\dfrac{\pi}{4}\right)+cot\cdot\left(2x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=-cot\cdot\left(2x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=cot\cdot\left(-2x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{2}+2x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{6}+2x\right)\)
\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+2x+k\pi\)
\(\Leftrightarrow-x=\dfrac{-\pi}{12}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{12}-k\pi\left(k\in Z\right)\)