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

1 Answer
Nghi N.
Aug 7, 2016

(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

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
1792 views around the world
You can reuse this answer
Creative Commons License