SOLUTION: Please solve, 4cos^2beta-3=0. The answers are pi/6,5pi/6,7pi/6,11pi/6 but i don't know how to get those answers
Algebra ->
Trigonometry-basics
-> SOLUTION: Please solve, 4cos^2beta-3=0. The answers are pi/6,5pi/6,7pi/6,11pi/6 but i don't know how to get those answers
Log On
Question 932566: Please solve, 4cos^2beta-3=0. The answers are pi/6,5pi/6,7pi/6,11pi/6 but i don't know how to get those answers Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! 4cos^2beta-3=0
cos^2beta = 3/4
cos beta = ± √3 / 2
See Below: beta = π/6, 5π/6, 7π/6, 11π/6
.......
(cosx, sinx) Summary Unit Circle .
.