Question

If `tan 2\beta = - \sqrt 3` and `0 <= \beta <= \pi`, what is `\beta` ?

Answer 1

`pi/3; (5pi)/6`

Explanation:

`tan 2t = - sqrt3`
Use trig identity:
`tan 2t = (2tan t)/(1 - tan^2 t)`
In this case:
`- sqrt3 = (2tan t)/(1 - tan^2 t)`
`-sqrt3 + sqrt3tan^2 t = 2tan t`
`sqrt3tan^2 t - 2tan t - sqrt3 = 0`
Solve this quadratic equation for tan t by using the improved quadratic formula (Socratic, Google Search):

`D = d^2 = b^2 - 4ac = 4 + 12 = 16` --> `d = +- 4`
There are 2 real roots:
`tan t = -b/(2a) +- d/(2a) = 2/(2sqrt3) +- 4/(2sqrt3) =`
`= (sqrt3+- 2sqrt3)/3`. Two solutions:
`tan t = sqrt3`, and `tan t = - sqrt3/3`
Inside the interval `[0, pi]` , trig table and unit circle give 2 answers:
`t = pi/3` and `t = (5pi)/6`