Restriction in the trigonometric identity of $\tan 3x$
Solution 1:
Suppose $x=n\pi+\pi/2.$ Then it follows that $3x=(3n+1)\pi+\pi/2.$ So the restriction on $x$ is already implied by the restriction on $3x.$
Suppose $x=n\pi+\pi/2.$ Then it follows that $3x=(3n+1)\pi+\pi/2.$ So the restriction on $x$ is already implied by the restriction on $3x.$