Question 810319
Solve the equation cos2theta=-√3/2
A) On the Interval [0,2pi)
B) On (-infinity,Infinity)
***
cos(2x)=1-2sin^2x=-√3/2
2sin^2x=1+√3/2=(2+√3)/2
sin^2x=(2+√3)/4
sinx=±√(2+√3)/4=±√(2+√3)/2=±0.9659

A)[0,2pi)
x=1.3089,1.8327,4.4504,4.9743
..
B)(-∞,∞)
1.3089±2πk,1.8327±2πk,4.4504±2πk,4.9743±2πk, k=any integer