How do you solve $tan 3 x (tan x - 1) = 0$ $tan3x(tanx-1)=0$ ?

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Answer 1 · 2016-05-23T09:57:08

$x = \frac{n π}{3}$ $x=(npi)/3$ or $x = n π + \frac{π}{4}$ $x=npi+pi/4$ , where $n$ $n$ is an integer.

As $tan 3 x (tan x - 1) = 0$ $tan3x(tanx-1)=0$

We have either $tan 3 x = 0$ $tan3x=0$ , which gives us

$3 x = n π$ $3x=npi$ , where $n$ $n$ is an integer i.e. $x = \frac{n π}{3}$ $x=(npi)/3$

or $tan x - 1 = 0$ $tanx-1=0$ i.e. $tan x = \frac{tan π}{4}$ $tanx=tanpi/4$ which gives us

$x = n π + \frac{π}{4}$ $x=npi+pi/4$

How do you solve tan3x(tanx-1)=0tan3x(tanx−1)=0tan3x(tanx-1)=0?