A alyren Junior Member Joined Sep 9, 2010 Messages 59 Nov 15, 2010 #1 solve equation tan 2x ? tan x = 0 [(2tanx)/(1-tan^2x)] ? tan x = 0 [(2tanx - tanx(1-tan^2x))/(1-tan^2x)] = 0 stuck here
solve equation tan 2x ? tan x = 0 [(2tanx)/(1-tan^2x)] ? tan x = 0 [(2tanx - tanx(1-tan^2x))/(1-tan^2x)] = 0 stuck here
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Nov 15, 2010 #2 Hello, alyren! \(\displaystyle \text{Solve: }\;\tan 2x - \tan x \:=\: 0\) Click to expand... \(\displaystyle \text{Note that }x\text{ cannot be a multiple of }\tfrac{\pi}{4}.\) \(\displaystyle \text{We have: }\;\frac{2\tan x}{1-\tan^2x} \:=\: \tan x\) \(\displaystyle \text{Multiply by }1 - \tan^2\!x\!:\;\; 2\tan x \:=\:\tan x - \tan^3x \quad\Rightarrow\quad \tan^3\!x + \tan x \:=\:0\) \(\displaystyle \text{Factor: }\:\tan x(\tan^2\!x+1) \:=\:0\) \(\displaystyle \text{Therefore: }\;\begin{Bmatrix}\tan x \:=\:0 & \Rightarrow & x \:=\:\pi n & \text{for }n \in I \\ \tan^2\!x+1\:=\:0 & \Rightarrow & \tan^2x \:=\:-1 & \text{no real roots} \end{Bmatrix}\)
Hello, alyren! \(\displaystyle \text{Solve: }\;\tan 2x - \tan x \:=\: 0\) Click to expand... \(\displaystyle \text{Note that }x\text{ cannot be a multiple of }\tfrac{\pi}{4}.\) \(\displaystyle \text{We have: }\;\frac{2\tan x}{1-\tan^2x} \:=\: \tan x\) \(\displaystyle \text{Multiply by }1 - \tan^2\!x\!:\;\; 2\tan x \:=\:\tan x - \tan^3x \quad\Rightarrow\quad \tan^3\!x + \tan x \:=\:0\) \(\displaystyle \text{Factor: }\:\tan x(\tan^2\!x+1) \:=\:0\) \(\displaystyle \text{Therefore: }\;\begin{Bmatrix}\tan x \:=\:0 & \Rightarrow & x \:=\:\pi n & \text{for }n \in I \\ \tan^2\!x+1\:=\:0 & \Rightarrow & \tan^2x \:=\:-1 & \text{no real roots} \end{Bmatrix}\)
