Question

Giải phương trình :

tanx/(1-tan^2x)=cot(x+pi/4)/2

Answer 1

Ta có \(\frac{\tan x}{1-\tan^2x}=\frac{\cot\left(x+\frac{\pi}{4}\right)}{2}\)

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\sin\left(x+\frac{\pi}{4}\right)\ne0\\cos\left(x+\frac{\pi}{4}\right)\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne k\pi-\frac{\pi}{4}\end{matrix}\right.\)

\(\Leftrightarrow2\cdot\frac{sinx.cos^2x}{cosx.\left(cos^2x-sin^2x\right)}=\frac{sinx+cosx}{cosx-sinx}\)

\(\Leftrightarrow2sinx.cosx=\left(sinx+cosx\right)^2=1+2sinx.cosx\)

=> PTVN