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\(DKXD:\left\{{}\begin{matrix}\cos\left(2x+\frac{\pi}{8}\right)\ne0\\\sin\left(x-\frac{3\pi}{4}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+\frac{\pi}{8}\ne\frac{\pi}{2}+k\pi\\x-\frac{3\pi}{4}\ne k\pi\end{matrix}\right.\)
\(pt\Leftrightarrow\tan\left(2x+\frac{\pi}{8}\right)=-\cot\left(x-\frac{3\pi}{4}\right)=\tan\left(x-\frac{3\pi}{4}+\frac{\pi}{2}\right)\)
\(\Leftrightarrow2x+\frac{\pi}{8}=x-\frac{3\pi}{4}+\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=-\frac{3}{8}\pi+k\pi\)
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) \(x=300^0+k540^0,k\in\mathbb{Z}\)
\(tan\cdot\left(x+\dfrac{\pi}{4}\right)+cot\cdot\left(2x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=-cot\cdot\left(2x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=cot\cdot\left(-2x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{2}+2x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{6}+2x\right)\)
\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+2x+k\pi\)
\(\Leftrightarrow-x=\dfrac{-\pi}{12}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{12}-k\pi\left(k\in Z\right)\)
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow tan^2x-2cot^2x+2=0\)
Đặt \(tan^2x=a>0\)
\(a-\frac{2}{a}+2=0\)
\(\Leftrightarrow a^2+2a-2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=\sqrt{3}-1\\a=-\sqrt{3}-1< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow tan^2x=\sqrt{3}-1\Rightarrow tanx=\pm\sqrt{\sqrt{3}-1}=tan\left(\pm\alpha\right)\)
\(\Rightarrow x=\pm\alpha+k\pi\)
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 \(cos\frac{2\pi}{5}\ne0\)
\(\Leftrightarrow\frac{cos3x}{sin3x}-\frac{sin\frac{2\pi}{5}}{cos\frac{2\pi}{5}}=0\)
\(\Leftrightarrow cos3x\cdot cos\frac{2\pi}{5}-sin\frac{2\pi}{5}\cdot sin3x=0\)
\(\Leftrightarrow cos\left(3x+\frac{2\pi}{5}\right)=0\)
\(\Leftrightarrow3x+\frac{2\pi}{5}=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{30}+\frac{k\pi}{3}\)
Ta có: \(\tan^2x+\cot^2x=2\)
\(\Leftrightarrow\tan^2x+2+\frac{1}{\tan^2x}=4\)
\(\Leftrightarrow\left(\tan x+\frac{1}{\tan x}\right)^2=4\)
\(\Leftrightarrow\left(\frac{\sin x}{\cos x}+\frac{1}{\frac{\sin x}{\cos x}}\right)^2=4\)
\(\Leftrightarrow\left(\frac{\sin^2x+\cos^2x}{\sin x.\cos x}\right)^2=4\)
\(\Leftrightarrow\left(\frac{1}{\sin x.\cos x}\right)^2=4\)
\(\Leftrightarrow4.\sin^2x.\cos^2x=1\)
\(\Leftrightarrow\sin^22x=1\)
\(\Leftrightarrow\orbr{\begin{cases}\sin2x=1\\\sin2x=-1\end{cases}}\Rightarrow2x=\left(2n-1\right)\cdot\frac{\pi}{2}\)
\(\Rightarrow x=\left(2n-1\right)\cdot\frac{\pi}{4}=\frac{n\pi}{2}-\frac{\pi}{4}\) (với n là số tự nhiên)