Mathematics - Trigonometric Equation Question with Solution | TestHub

The equation $\tan x+\tan 2x=\tan 3x$ can be written as $\tan x+\tan 2x-\tan (x+2x)=0$. Using the identity $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ if $A+B+C=n\pi$, since $x+2x+(-3x)=0$, the equation simplifies to $\tan x\tan 2x\tan 3x=0$. This implies $\tan x=0$ or $\tan 2x=0$ or $\tan 3x=0$. Thus, $x=n\pi /k$ for $k\in \{1,2,3\}$, which means $x=n\pi /3$. However, we must ensure that $\tan x$, $\tan 2x$, and $\tan 3x$ are defined, i.e., $\cos x\ne 0$, $\cos 2x\ne 0$, and $\cos 3x\ne 0$. In the interval $[0,2\pi ]$, the values $x=n\pi /3$ are $0,\pi /3,2\pi /3,\pi ,4\pi /3,5\pi /3,2\pi$. We must exclude values that make any denominator zero. For instance, $x=\pi /2$ and $x=3\pi /2$ are excluded because $\cos (3x)$ becomes zero (e.g., $3(\pi /2)=3\pi /2$). All other values $0,\pi /3,2\pi /3,\pi ,4\pi /3,5\pi /3,2\pi$ are valid solutions. Counting these, there are 7 solutions.