The general solution of the trigonometric equ | vedclass

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(D) Given equation: $	an x + 	an 2x + 	an 3x = 	an x 	an 2x 	an 3x$.We know that $	an(A+B+C) = \frac{	an A + 	an B + 	an C - 	an A 	an B \

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$x = \frac{n\pi}{3}$,where $n \in I$

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(D) Given equation: $	an x + 	an 2x + 	an 3x = 	an x 	an 2x 	an 3x$.We know that $	an(A+B+C) = \frac{	an A + 	an B + 	an C - 	an A 	an B 	an C}{1 - (	an A 	an B + 	an B 	an C + 	an C 	an A)}$.If $	an A + 	an B + 	an C = 	an A 	an B 	an C$,then $	an(A+B+C) = 0$.Here,$A=x, B=2x, C=3x$,so $A+B+C = 6x$.Thus,$	an(6x) = 0$.This implies $6x = n\pi$,where $n \in I$.Therefore,$x = \frac{n\pi}{6}$.However,we must check for domain restrictions where $	an x, 	an 2x, 	an 3x$ are defined.For $	an 3x$ to be defined,$3x 
eq (2k+1)\frac{\pi}{2} \Rightarrow x 
eq (2k+1)\frac{\pi}{6}$.Given the options provided,the general solution is $x = \frac{n\pi}{3}$.

The general solution of the trigonometric equation $\tan x + \tan 2x + \tan 3x = \tan x \cdot \tan 2x \cdot \tan 3x$ is

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