Trigonometric Equations

The foundational step is to apply the general solution of an equality of tangents, which states that whenever $\tan A=\tan B$ the angles differ by an integer multiple of $\pi$, that is $A=B+n\pi$ for some integer $n$. This is because tangent has period $\pi$, so equal tangents need not have equal angles, only angles separated by a half-turn. Setting $A=3x$ and $B=x$ gives $3x=x+n\pi$, hence $2x=n\pi$ and the candidate family $x=\frac{n\pi }{2}$. The crucial subtlety, which separates a correct JEE Advanced answer from a careless one, is that the original equation only makes sense where both $\tan 3x$ and $\tan x$ are defined, so any candidate making either tangent blow up must be discarded. When $n$ is odd, $x=\frac{n\pi }{2}$ is an odd multiple of $\frac{\pi }{2}$, exactly where $\tan x$ is undefined, so those values fail. Only even $n$ survive, and writing $n=2m$ gives $x=m\pi$, equivalently $x=n\pi$ for integer $n$. Option $\frac{n\pi }{2}$ ignores this domain restriction and retains invalid points. Option $\frac{n\pi }{3}$ stems from a coefficient slip. Option $(2n+1)\frac{\pi }{2}$ selects precisely the excluded undefined points. This reflects the standard tangent-equation domain-checking pattern. As a final verification, at $x=\pi$ both $\tan 3\pi =0$ and $\tan \pi =0$ are defined and equal, while at $x=\frac{\pi }{2}$ the tangent is undefined and is correctly excluded.