How do you solve $3 {cot}^{2} x - 1 = 0$ $3cot^2x-1=0$ between the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

Question

21924 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-14T04:54:44

In $(0, 2 π), x = \frac{π}{6}, 5 \frac{π}{6}, 7 \frac{π}{6} and 11 \frac{π}{6}$ $(0, 2pi), x =pi/6, 5pi/6, 7pi/6 and 11pi/6$

Here, ${tan}^{2} x = \frac{1}{3}$ $tan^2 x=1/3$ . So,

$tan x = \pm \frac{1}{\sqrt{3}}$ $tanx =+-1/sqrt3$ .

The principal values for x are $\pm \frac{π}{6}$ $+-pi/6$ and the genera values are

$npi+-pi/6, n =0, +-1, +-2, +-3, ...$

$=+-pi/6, +-5pi/6, +-7pi/6 ...$

In $(0, 2pi), x =pi/6, 5pi/6, 7pi/6 and 11pi/6$ .

How do you solve 3cot^2x-1=03cot2x−1=03cot^2x-1=0 between the interval 0<=x<=2pi0≤x≤2π0<=x<=2pi?