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April 19th, 2018, 07:22 AM   #1
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Question General Solution

Find General Solution

Tan 2theta = - Tan theta
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April 19th, 2018, 08:05 AM   #2
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$\tan(2\theta) = -\tan(\theta)$

$\dfrac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta)-\sin^2(\theta)} = -\dfrac{\sin(\theta)}{\cos(\theta)}$

$2\sin(\theta)\cos^2(\theta) = -\sin(\theta)\cos^2(\theta)+\sin^3(\theta)$

$3\sin(\theta)\cos^2(\theta) = \sin^3(\theta)$

$3\sin(\theta)(1-\sin^2(\theta)) = \sin^3(\theta)$

$3\sin(\theta) - 3\sin^3(\theta) = \sin^3(\theta)$

$3\sin(\theta) = 4\sin^3(\theta)$

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.

$\theta = k\pi,~k \in \mathbb{Z}$
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April 19th, 2018, 08:12 AM   #3
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Thank you, but book gives answer of

Theta is 60
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April 19th, 2018, 08:22 AM   #4
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Quote:
Originally Posted by Vee88 View Post
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 View Post
Thank you, but book gives answer of

Theta is 60
Quote:
Originally Posted by romsek View Post
$\tan(2\theta) = -\tan(\theta)$
$\theta = \dfrac{\pi}{3}+2k\pi,~\dfrac{2\pi}{3}+2k\pi,~ \dfrac{4\pi}{3}+2k\pi,~\dfrac{5\pi}{3}+2k\pi,~k \in \mathbb{Z}$

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
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April 19th, 2018, 11:44 PM   #6
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Thank you so much
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April 20th, 2018, 01:13 AM   #7
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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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