Cot(θ) = tan(2θ - 3π) find 0 < θ < 2π

[karush]

$\cot{(\theta)}=\tan{(2\theta-3\pi)}$ find $0<\theta<2\pi$

From the periodic Formula $\tan{(\theta+\pi n)}=\tan{\theta}$

thus
$
\displaystyle
\cot{(\theta)}
=\tan{(2\theta)}
\Rightarrow
\frac{1}{\tan{\theta}}
=\tan{(2\theta)}
$

there are 6 answers to this, but stuck here

 

[MarkFL]

I would stop at:

$$\cot(\theta)=\tan(2\theta)$$

Then, I would try combining the following:

Co-function identity:

$$\cot(\theta)=\tan\left(\frac{\pi}{2}-\theta \right)$$

Periodicity of tangent function:

$$\tan(\theta)=\tan(\theta+k\pi)$$ where $$k\in\mathbb{Z}$$

This will give you the six roots you desire.

 

[karush]

by this I assume

$\displaystyle 2 \theta =\frac{\pi}{2}-\theta$
so
$\displaystyle
\theta=\frac{\pi}{6}$
then for $\displaystyle 0<\theta<2\pi $ using $\tan{(\theta+\pi n)}=\tan{\theta}$ for period
$$
\theta = \frac{\pi}{6}, \frac{7\pi}{6}
$$
or
$$30^o,210^o$$

but that is only 2 of them

 

[MarkFL]

You would actually have:

$$\theta=\frac{\pi}{2}-2\theta+k\pi$$

See what you get from that. :D

 

[soroban]

Hello, karush!

\cot \theta\:=\:\tan(2\theta-3\pi),\quad 0&lt;\theta&lt;2\pi

\tan(2\theta - 3\pi) \;=\;\frac{\tan(2\theta) - \tan(3\pi)}{1 + \tan(2\theta)\tan(3\pi)} \;=\;\tan(2\theta)

. . . . . . . . . . =\;\frac{2\tan\theta}{1-\tan^2\theta}

The equation becomes: .\frac{1}{\tan\theta} \;=\;\frac{2\tan\theta}{1-\tan^2\theta}

. . 1 - \tan^2\theta \:=\:2\tan^2\theta \quad\Rightarrow\quad 3\tan^2\theta \:=\:1

. . \tan^2\theta \:=\:\frac{1}{3} \quad\Rightarrow\quad \tan\theta \:=\:\pm\frac{1}{\sqrt{3}}

Therefore: .\theta \;=\;\frac{\pi}{6},\;\frac{5\pi}{6},\;\frac{7\pi}{6},\;\frac{11\pi}{6}