How do you solve #sinx=-sqrt3/2# on the interval [0,2pi]?
1 Answer
Feb 20, 2016
Explanation:
Since the ratio has a negative value , this informs us that the required solutions will be in the 3rd and 4th quadrants. The sin ratio is positive in the 1st/2nd quadrants.
now
#sin^-1(-sqrt3/2) = -pi/3 #
# rArr x = (pi + pi/3 ) =( 4pi)/3 , x = (2pi - pi/3 ) =( 5pi)/3 #
