Question

How do you solve `2sin((3x)/2)+sqrt3=0`?

Answer 1

`(8pi)/9 + 2kpi`
`(10pi)/9 + 2kpi`

Explanation:

`2sin ((3x)/2) + sqrt3 = 0`
`sin ((3x)/2) = - sqrt3/2`
Trig table of special arcs, and unit circle give 2 arcs (3x)/2:
arc `(3x)/2 = - (2pi)/3` , and arc `(3x)/2 = pi - ((-2pi)/3) = (5pi)/3`
a. Arc `(-2pi)/3` --> is the same as arc `(4pi)/3` (co-terminal)
`(3x)/2 = (4pi)/3` -->`x = (4pi/3)(2/3) = (8pi)/9`

b. `(3x)/2 = (5pi)/3` --> `x = (5pi/3)(2/3) = (10pi)/9`
General answers:
`x = (8pi)/9 + 2kpi`
`x = (10pi)/9 + 2kpi`