# Tan(2x)-cot(x)=0?

## I solved it and got the solutions: pi/6, 5pi/6, 7pi/6 and 11pi/6. Why did I not get the pi/2 and 3pi/2 solutions?

Feb 18, 2018

My solution is found in explanation section.

#### Explanation:

$\tan 2 x - \cot x = 0$

$\tan 2 x = \cot x$

$\tan 2 x \cdot \tan x = 1$

$\frac{\sin 2 x \cdot \sin x}{\cos 2 x \cdot \cos x} = 1$

$\frac{\frac{1}{2} \cdot \left(\cos x - \cos 3 x\right)}{\frac{1}{2} \cdot \left(\cos 3 + \cos x\right)} = 1$

$\frac{\cos x - \cos 3 x}{\cos 3 x + \cos x} = 1$

$\cos x - \cos 3 x = \cos 3 x + \cos x$

$2 \cos 3 x = 0$

$\cos 3 x = 0$

I have $2$ states for $\cos 3 x = 0$,

a) $\cos 3 x = \cos \left(\frac{\pi}{2} + 2 \pi \cdot k\right)$

$3 x = \frac{\pi}{2} + 2 \pi \cdot k$, so $x = \frac{\pi}{6} + \frac{2 \pi}{3} \cdot k$

Consequently, ${x}_{1} = \frac{\pi}{6}$, ${x}_{2} = \frac{5 \pi}{6}$ and ${x}_{3} = \frac{3 \pi}{2}$ for $k = 0 , 1$ and $2$

b) $\cos 3 x = \cos \left(\frac{3 \pi}{2} + 2 \pi \cdot k\right)$

$3 x = \frac{3 \pi}{2} + 2 \pi \cdot k$, so $x = \frac{\pi}{2} + \frac{2 \pi}{3} \cdot k$

Consequently, ${x}_{4} = \frac{\pi}{2}$, ${x}_{5} = \frac{7 \pi}{6}$ and ${x}_{6} = \frac{11 \pi}{6}$ for $k = 0 , 1$ and $2$