Question

Sin x > -√3/2 please Solve in R inequation:

Answer 1

(4π/3+2πn; 5π/3+2πn)

Explanation:

1. to calculate the zeros:

sinx=-sqrt(3)/2; ⇔

`x=(-1)^(n+1)*((\pi )/(3))+\pi *n`, where n∈Z.

2. using the zeros found and the attached picture:

x∈(4π/3+2πn; 5π/3+2πn).