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 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: 2,311 Thanks: 1224 $\tan(2\theta) = -\tan(\theta)$ $\dfrac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta)-\sin^2(\theta)} = -\dfrac{\sin(\theta)}{\cos(\theta)}$ $2\sin(\theta)\cos^2(\theta) = -\sin(\theta)\cos^2(\theta)+\sin^3(\theta)$ $3\sin(\theta)\cos^2(\theta) = \sin^3(\theta)$ $3\sin(\theta)(1-\sin^2(\theta)) = \sin^3(\theta)$ $3\sin(\theta) - 3\sin^3(\theta) = \sin^3(\theta)$ $3\sin(\theta) = 4\sin^3(\theta)$ 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. $\theta = k\pi,~k \in \mathbb{Z}$ Thanks from topsquark
 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
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Quote:
 Originally Posted by Vee88 Thank you, but book gives answer of Theta is 60
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.
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April 19th, 2018, 01:33 PM   #5
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Quote:
 Originally Posted by Vee88 Thank you, but book gives answer of Theta is 60
Quote:
 Originally Posted by romsek $\tan(2\theta) = -\tan(\theta)$ $\theta = \dfrac{\pi}{3}+2k\pi,~\dfrac{2\pi}{3}+2k\pi,~ \dfrac{4\pi}{3}+2k\pi,~\dfrac{5\pi}{3}+2k\pi,~k \in \mathbb{Z}$ Additionally solutions occur where $\sin(\theta)=0$, i.e. $\theta = k\pi,~k \in \mathbb{Z}$
That would be the first of romsek's solutions for k = 0, $\displaystyle \theta = \frac{\pi}{3} + 2(0) \pi = \frac{\pi}{3}$

-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: 20,285 Thanks: 1968 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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