Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(\tan^2x-\cot^2\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-1-\frac{1}{\sin^2\left(x-\frac{\pi}{4}\right)}+1=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\sin x.\cos\frac{\pi}{4}-\cos x.\sin\frac{\pi}{4}\right)^2}=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x\right)^2}=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x}=0\)
\(\Leftrightarrow\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x-\cos^2x=0\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}\cos^2x-\sin x.\cos x-\frac{1}{2}\cos^2x=0\)
\(\Leftrightarrow\cos^2x+\sin x.\cos x-\frac{1}{2}=0\)
Đến đây là dễ r nha bn :3
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.
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.
b)đề là \(tan\left(x-15^0\right)=\frac{\sqrt{3}}{3}\)
Vì \(\frac{\sqrt{3}}{3}=tan30^0\) nên
\(\Leftrightarrow tan\left(x-15^0\right)=tan30^0\)
\(\Leftrightarrow x-15^0=30^0+k180^0\)
\(\Leftrightarrow x=45^0+k180^0\left(k\in Z\right)\)
Đk:\(sin3x\ne0\) và \(cos\frac{2\pi}{5}\ne0\)
\(\Leftrightarrow\frac{cos3x}{sin3x}-\frac{sin\frac{2\pi}{5}}{cos\frac{2\pi}{5}}=0\)
\(\Leftrightarrow cos3x\cdot cos\frac{2\pi}{5}-sin\frac{2\pi}{5}\cdot sin3x=0\)
\(\Leftrightarrow cos\left(3x+\frac{2\pi}{5}\right)=0\)
\(\Leftrightarrow3x+\frac{2\pi}{5}=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{30}+\frac{k\pi}{3}\)
a: tan x=1/căn 3
=>tan x=tan(pi/6)
=>x=pi/6+kpi
b: tan(30-3x)=tan75
=>30-3x=75+k*180
=>3x=-45-k*180
=>x=-15-k*60
c: \(cot3x=cot\left(\dfrac{3}{4}pi\right)\)
=>3x=3/4pi+kpi
=>x=1/4pi+kpi/3
d: cot(5x+30 độ)=cot 75 độ
=>5x+30=75+k*180
=>5x=45+k*180
=>x=9+k*36
a) \(\sqrt 3 \tan 2x = - 1\;\; \Leftrightarrow \tan 2x = - \frac{1}{{\sqrt 3 }}\;\;\; \Leftrightarrow \tan 2x = \tan - \frac{\pi }{6}\; \Leftrightarrow 2x = - \frac{\pi }{6} + k\pi \)
\(\;\; \Leftrightarrow x = - \frac{\pi }{{12}} + \frac{{k\pi }}{2}\;\left( {k \in \mathbb{Z}} \right)\)
b) \(\tan 3x + \tan 5x = 0\;\; \Leftrightarrow \tan 3x = \tan \left( { - 5x} \right) \Leftrightarrow 3x = - 5x + k\pi \;\; \Leftrightarrow 8x = k\pi \;\; \Leftrightarrow x = \frac{{k\pi }}{8}\;\left( {k \in \mathbb{Z}} \right)\)
a: \(PT\Leftrightarrow tan\left(2x-30^0\right)=-\sqrt{3}\)
=>\(2x-30^0=-60^0+k\cdot180^0\)
=>\(2x=-30^0+k\cdot180^0\)
=>\(x=-15^0+k\cdot90^0\)
b: \(cot2x-1=0\)
=>cot2x=1
=>\(2x=\dfrac{\Omega}{4}+k\cdot\Omega\)
=>\(x=\dfrac{\Omega}{8}+\dfrac{k\Omega}{2}\)
c: \(cot3x+\sqrt{3}=0\)
=>\(cot3x=-\sqrt{3}\)
=>\(3x=-\dfrac{\Omega}{6}+k\Omega\)
=>\(x=-\dfrac{\Omega}{18}+\dfrac{k\Omega}{3}\)