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a: \(y=3\cdot tanx-2\cdot cotx\)

=>\(y'=3\left(tan^2x+1\right)-2\cdot\left(-1\right)\cdot\left(cot^2x+1\right)\)

=>\(y'=3\cdot tan^2x+3+2\cdot cot^2x+2\)

=>\(y'=3\cdot tan^2x+2\cdot cot^2x+5\)

b: \(y=3\cdot cot\left(2x+\dfrac{\Omega}{3}\right)\)

=>\(y'=3\cdot\left(-1\right)\cdot\left(2x+\dfrac{\Omega}{3}\right)'\cdot\left(cot^2\left(2x+\dfrac{\Omega}{3}\right)+1\right)\)

=>\(y'=-6\cdot\left(cot^2\left(2x+\dfrac{\Omega}{3}\right)+1\right)\)

=>\(y'=-6\cdot cot^2\left(2x+\dfrac{\Omega}{3}\right)-6\)

c: \(y=tan\left(3x-4\right)+cot\left(x-5\right)\)

=>\(y'=\left(3x-4\right)'\cdot\left(tan^2\left(3x-4\right)+1\right)+\left(-1\right)\cdot\left(x-5\right)'\cdot\left(cot^2\left(x-5\right)+1\right)\)

=>\(y'=3\cdot\left(tan^2\left(3x-4\right)+1\right)-1\cdot\left(cot^2\left(x-5\right)+1\right)\)

=>\(y'=3\cdot tan^2\left(3x-4\right)-cot^2\left(x-5\right)+2\)
d: \(y=5\cdot tan\left(3x-\dfrac{\Omega}{5}\right)-3\cdot cot\left(2x+\dfrac{7}{6}\Omega\right)\)
=>\(y'=5\cdot\left(3x-\dfrac{\Omega}{5}\right)'\cdot\left(tan^2\left(3x-\dfrac{\Omega}{5}\right)+1\right)+3\cdot\left(2x+\dfrac{7}{6}\Omega\right)'\cdot\left(cot^2\left(2x+\dfrac{7}{6}\Omega\right)+1\right)\)

=>\(y'=15\left(tan^2\left(3x-\dfrac{\Omega}{5}\right)+1\right)+6\left(cot^2\left(2x+\dfrac{7}{6}\Omega\right)+1\right)\)

=>\(y'=15\cdot tan^2\left(3x-\dfrac{\Omega}{5}\right)+6\cdot cot^2\left(2x+\dfrac{7}{6}\Omega\right)+21\)

e: \(y=3\cdot tan^3x-cot^43x\)

=>\(y'=3\cdot3\cdot tan^2x\cdot\left(tanx\right)'-4\cdot cot^33x\cdot\left(cot3x\right)'\)

=>\(y'=9tan^2x\cdot\left(tan^2x+1\right)-4\cdot cot^33x\cdot\left(-1\right)\cdot\left(3x\right)'\cdot\left(cot^23x+1\right)\)

=>\(y'=9\cdot tan^4x+9tan^2x+12\cdot cot^33x\left(cot^23x+1\right)\)

=>\(y'=9\cdot tan^4x+9\cdot tan^2x+12\cdot cot^53x+12\cdot cot^33x\)

NV
15 tháng 7 2021

a.

\(\left\{{}\begin{matrix}sin\left(3x+\dfrac{\pi}{6}\right)\ne0\\cos2x\ne0\\sinx\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

b.

Do \(5+2cot^2x-sinx=4+2cot^2x+\left(1-sinx\right)>0\) nên hàm xác định khi:

\(\left\{{}\begin{matrix}sinx\ne0\\sin\left(x+\dfrac{\pi}{2}\right)\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow sin2x\ne0\)

\(\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

6 tháng 8 2021

a, y xác định `<=> 3cos(2x+3) \ne 0`

`<=>cos(2x+3) \ne 0`

`<=>2x+3 \ne π/2+kπ`

`<=>x \ne π/4 -3/2 +k π/2 (k \in ZZ)`

b, y xác định `<=> sin(x/3+π/4) \ne0`

`<=> x/3+π/4 \ne kπ`

`<=> x \ne (-3π)/4+ k3π`

NV
6 tháng 8 2021

ĐKXĐ: 

a.

\(cos\left(2x+3\right)\ne0\)

\(\Leftrightarrow2x+3\ne\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=-\dfrac{3}{2}+\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

b.

\(sin\left(\dfrac{x}{3}+\dfrac{\pi}{4}\right)\ne0\)

\(\Leftrightarrow\dfrac{x}{3}+\dfrac{\pi}{4}\ne k\pi\)

\(\Leftrightarrow x\ne-\dfrac{3\pi}{4}+k3\pi\)

18 tháng 5 2017

Hàm số lượng giác, phương trình lượng giác

Hàm số lượng giác, phương trình lượng giác

loading...  loading...  

a: ĐKXĐ: 2*sin x+1<>0

=>sin x<>-1/2

=>x<>-pi/6+k2pi và x<>7/6pi+k2pi

b: ĐKXĐ: \(\dfrac{1+cosx}{2-cosx}>=0\)

mà 1+cosx>=0

nên 2-cosx>=0

=>cosx<=2(luôn đúng)

c ĐKXĐ: tan x>0

=>kpi<x<pi/2+kpi

d: ĐKXĐ: \(2\cdot cos\left(x-\dfrac{pi}{4}\right)-1< >0\)

=>cos(x-pi/4)<>1/2

=>x-pi/4<>pi/3+k2pi và x-pi/4<>-pi/3+k2pi

=>x<>7/12pi+k2pi và x<>-pi/12+k2pi

e: ĐKXĐ: x-pi/3<>pi/2+kpi và x+pi/4<>kpi

=>x<>5/6pi+kpi và x<>kpi-pi/4

f: ĐKXĐ: cos^2x-sin^2x<>0

=>cos2x<>0

=>2x<>pi/2+kpi

=>x<>pi/4+kpi/2

 

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...