April 19th, 2018, 07:22 AM  #1 
Newbie Joined: Apr 2018 From: East London Posts: 12 Thanks: 0  General Solution
Find General Solution Tan 2theta =  Tan theta 
April 19th, 2018, 08:05 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,165 Thanks: 1139 
$\tan(2\theta) = \tan(\theta)$ $\dfrac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta)\sin^2(\theta)} = \dfrac{\sin(\theta)}{\cos(\theta)}$ $2\sin(\theta)\cos^2(\theta) = \sin(\theta)\cos^2(\theta)+\sin^3(\theta)$ $3\sin(\theta)\cos^2(\theta) = \sin^3(\theta)$ $3\sin(\theta)(1\sin^2(\theta)) = \sin^3(\theta)$ $3\sin(\theta)  3\sin^3(\theta) = \sin^3(\theta)$ $3\sin(\theta) = 4\sin^3(\theta)$ if $\sin(\theta) \neq 0$ $3 = 4\sin^2(\theta)$ $\sin^2(\theta) = \dfrac 3 4$ $\sin(\theta) = \pm \dfrac{\sqrt{3}}{2}$ $\theta = \dfrac{\pi}{3}+2k\pi,~\dfrac{2\pi}{3}+2k\pi,~ \dfrac{4\pi}{3}+2k\pi,~\dfrac{5\pi}{3}+2k\pi,~k \in \mathbb{Z}$ Additionally solutions occur where $\sin(\theta)=0$, i.e. $\theta = k\pi,~k \in \mathbb{Z}$ 
April 19th, 2018, 08:12 AM  #3 
Newbie Joined: Apr 2018 From: East London Posts: 12 Thanks: 0 
Thank you, but book gives answer of Theta is 60 
April 19th, 2018, 08:22 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,165 Thanks: 1139  well that's in degrees for a start I don't expect you to understand the Mathematica code but all the TRUEs at the bottom show that each of the values in the list called solutions is in fact a solution to your original trig equality. 
April 19th, 2018, 01:33 PM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,907 Thanks: 773 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
April 19th, 2018, 11:44 PM  #6 
Newbie Joined: Apr 2018 From: East London Posts: 12 Thanks: 0 
Thank you so much

April 20th, 2018, 01:13 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,865 Thanks: 1833 
As $\tan(2\theta) = \tan(\theta) = \tan(\theta)$, $2\theta = \theta + 180\text{k}^\circ$, where $\text{k}$ is an integer, and so $\theta = 60\text{k}^\circ$. 

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