Question

How do you solve `sinx=-sqrt3/2` on the interval [0,2pi]?

Answer 1

`x =( 4pi)/3 , (5pi)/3`

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`