Register Now

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Toán Lớp 11: cot($\frac{\pi}{2}$-3x)=0 cot(x+2)=3 cot(x-$\frac{\pi}{6}$)=$\frac{1}{\sqrt[]{3}}$ cot($\frac{x}{2}$-$\frac{\pi}{4}$)=1

Home/ Toán Lớp 11: cot($\frac{\pi}{2}$-3x)=0 cot(x+2)=3 cot(x-$\frac{\pi}{6}$)=$\frac{1}{\sqrt[]{3}}$ cot($\frac{x}{2}$-$\frac{\pi}{4}$)=1

Toán Lớp 11: cot($\frac{\pi}{2}$-3x)=0 cot(x+2)=3 cot(x-$\frac{\pi}{6}$)=$\frac{1}{\sqrt[]{3}}$ cot($\frac{x}{2}$-$\frac{\pi}{4}$)=1

Diễm Phúc Tháng Chín 17, 2022 Toán học 1 Comment 0 views

Toán Lớp 11: cot($\frac{\pi}{2}$-3x)=0
cot(x+2)=3
cot(x-$\frac{\pi}{6}$)=$\frac{1}{\sqrt[]{3}}$
cot($\frac{x}{2}$-$\frac{\pi}{4}$)=1

Comments ( 1 )

kimoanh34
17 Tháng Chín, 2022 at 8:04 chiều

Reply

$cot(\dfrac{\pi}{2}-3x)=0$

Điều kiện $sinx\ne0$

$⇔x\ne k\pi$

$⇔\dfrac{\pi}{2}-3x=\dfrac{\pi}{2}+k\pi$

$⇔x=k3\pi(k∈Z)(l)$

$cot(x+2)=3$

Điều kiện $sin(x+2)\ne0$

$⇔x+2\ne k\pi$

$⇔x\ne-2+ k\pi$

$⇔x+2=arccot(3)+k\pi$

$⇔x=-2+arccot(3)+k\pi$

$cot(x-\dfrac{\pi}{6})=\dfrac{1}{\sqrt[]{3}}=cot(\dfrac{\pi}{3})$

Điều kiện $sin(x-\dfrac{\pi}{6})\ne0$

$⇔x-\dfrac{\pi}{6}\ne k\pi$

$⇔x\ne \dfrac{\pi}{6}+k\pi$

$⇔x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi$

$⇔x=\dfrac{\pi}{2}+k\pi(k∈Z)$

Leave a reply

About Diễm Phúc