First slide

Trigonometric equations

Question

If the general solution of the trigonometric equation $\tan 3 x + \tan 2 x + \tan x = \tan 3 x . \tan 2 x . \tan x$ is $\frac{n π}{k}$ then k =

Difficult

Solution

$\begin{array}{l} W e h a v e 3 x - 2 x - x = 0 \\ \Rightarrow \tan 3 x + \tan (- 2 x) + \tan (- x) = \tan 3 x . \tan (- 2 x) . \tan (- x) \\ \Rightarrow \tan 3 x - \tan 2 x - \tan x = \tan 3 x . \tan 2 x . \tan x \\ ∴ G i v e n e q u a t i o n b e c o m e s \tan 3 x + \tan 2 x + \tan x = \tan 3 x - \tan 2 x - \tan x \\ \Rightarrow \tan 2 x + \tan x = 0 \Rightarrow \tan 2 x = \tan (- x) \\ \Rightarrow 2 x = n π - x \\ \Rightarrow 3 x = n π \\ \Rightarrow x = \frac{n π}{3}, n \in z \end{array}$

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