Solve: tan2x=tanx
tan2x=tanx
i.e. 2tanx1−tan2x=tanx
i.e. 2tanx=tanx−tan3x
i.e. tan3x+tanx=0
i.e. tanx(tan2x+1)=0
i.e. tanx=0 or 1+tan2x=0
But 1+tan2x=0 is not possible ∵ \tan^2 x \neq -1
Now, \tan x = 0 \implies x = n \pi, \;\;\; n \in Z