SOLUTION: solve equations (solve over the interval [0,2pi)) sin2x= root2/2

Algebra ->  Trigonometry-basics -> SOLUTION: solve equations (solve over the interval [0,2pi)) sin2x= root2/2      Log On


   

Question 1023509: solve equations (solve over the interval [0,2pi))
sin2x= root2/2
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
solve equations (solve over the interval [0,2pi))
sin2x= root2/2
-----------
sin%282x%29+=+sqrt%282%29%2F2
---
2x = pi/4, 3pi/4, 9pi/4, 11pi/4
x = 1/2 of those.