Question

tan x = -tan \(\frac{\Pi}{7}\)

tan (\(x^2+1\))=0

cot x = 3 tanx

Answer 1

\(tanx=-tan\frac{\pi}{7}\Leftrightarrow tanx=tan\left(-\frac{\pi}{7}\right)\)

\(\Leftrightarrow x=-\frac{\pi}{7}+k\pi\)

\(tan\left(x^2+1\right)=0\Leftrightarrow x^2+1=k\pi\) (\(k>0\))

\(\Leftrightarrow x=\pm\sqrt{k\pi-1}\)

\(cotx=3tanx\Leftrightarrow\frac{1}{tanx}=3tanx\Leftrightarrow tan^2x=\frac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=-\frac{1}{\sqrt{3}}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{\pi}{6}+k\pi\)