Mathematics - Trigonometric Equation Question with Solution | TestHub

The given equation is $\tan x+\tan 2x+\tan 3x=0$. We can rewrite this as $\tan x+\tan 2x=-\tan 3x$. Using the identity $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$, we have $\frac{\tan x+\tan 2x}{1-\tan x\tan 2x}=\tan (x+2x)=\tan 3x$. So, $\tan 3x(1-\tan x\tan 2x)=\tan x+\tan 2x$. Substituting this back into the equation: $\tan 3x(1-\tan x\tan 2x)=-\tan 3x$.$\tan 3x(1-\tan x\tan 2x+1)=0⟹\tan 3x(2-\tan x\tan 2x)=0$. Case 1: $\tan 3x=0$. This gives $3x=n\pi ⟹x=n\pi /3$, where $n\in \mathbb{Z}$. We must ensure $\tan x,\tan 2x,\tan 3x$ are defined. For $x=n\pi /3$, none of these tangents are undefined. Case 2: $2-\tan x\tan 2x=0$. This implies $\tan x\tan 2x=2$. Using $\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}$: $\tan x\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)=2$.$\frac{2{\tan }^{2}x}{1-{\tan }^{2}x}=2⟹2{\tan }^{2}x=2(1-{\tan }^{2}x)⟹{\tan }^{2}x=1-{\tan }^{2}x$.$2{\tan }^{2}x=1⟹{\tan }^{2}x=1/2⟹\tan x=\pm 1/\sqrt{2}$. This gives $x=n\pi \pm \arctan (1/\sqrt{2})$, where $n\in \mathbb{Z}$. These values do not make any of the original tangents undefined. Combining both cases, the general solution is $x=n\pi /3$ or $x=n\pi \pm \arctan (1/\sqrt{2})$, where $n\in \mathbb{Z}$.