The general solution of the equation tan x + | vedclass

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(A) Given equation: $	an x + 	an 2x - 	an 3x = 0$ We know that $	an 3x = 	an(x + 2x) = \frac{	an x + 	an 2x}{1 - 	an x 	an 2x}$ So,$	an x +

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$\left\{x \mid x = n\pi 	ext{ or } x = \frac{n\pi}{3}, n \in Zight\}$

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(A) Given equation: $	an x + 	an 2x - 	an 3x = 0$ We know that $	an 3x = 	an(x + 2x) = \frac{	an x + 	an 2x}{1 - 	an x 	an 2x}$ So,$	an x + 	an 2x = 	an 3x(1 - 	an x 	an 2x)$ $	an x + 	an 2x = 	an 3x - 	an x 	an 2x 	an 3x$ $	an x + 	an 2x - 	an 3x = -	an x 	an 2x 	an 3x$ Since the given equation is $	an x + 	an 2x - 	an 3x = 0$,we have: $-	an x 	an 2x 	an 3x = 0$ This implies $	an x = 0$ or $	an 2x = 0$ or $	an 3x = 0$ For $	an x = 0$,$x = n\pi$ For $	an 2x = 0$,$2x = n\pi \Rightarrow x = \frac{n\pi}{2}$ For $	an 3x = 0$,$3x = n\pi \Rightarrow x = \frac{n\pi}{3}$ Combining these,the set of solutions is $\left\{x \mid x = \frac{n\pi}{3}, n \in Zight\}$.

The general solution of the equation $\tan x + \tan 2x - \tan 3x = 0$ is

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