Question

Question #e3a44

Answer 1

Solution: In interval `[0,2pi], x= (4pi)/3, x= (5pi)/3`

Explanation:

`sinx = - sqrt3/2 ; sin (-pi/3)= - sqrt3/2` or

`-pi/3 = (2pi-pi/3) = (5pi)/3 :. sin((5pi)/3)= - sqrt3/2`

`:. x= (5pi)/3` In 3rd quadrant `sin theta` is negative

`:. sin(pi+pi/3)=- sqrt3/2 :.sin((4pi)/3)=- sqrt3/2:. x= (4pi)/3`

Solution: In interval `[0,2pi], x= (4pi)/3, x= (5pi)/3` [Ans]