SOLUTION: Can you help me solve this please? The question says, "find solutions between [0, 2PI)" The problem is sin(2theta)=sqrt(2)cos(theta) Thank You!!!!

Algebra ->  Trigonometry-basics -> SOLUTION: Can you help me solve this please? The question says, "find solutions between [0, 2PI)" The problem is sin(2theta)=sqrt(2)cos(theta) Thank You!!!!      Log On


   

Question 769266: Can you help me solve this please?
The question says, "find solutions between [0, 2PI)"
The problem is
sin(2theta)=sqrt(2)cos(theta)
Thank You!!!!
Found 2 solutions by lwsshak3, Alan3354:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
find solutions between [0, 2PI)"
sin(2theta)=sqrt(2)cos(theta)
sin2x=√2 cosx
2sinxcosx=√2cosx
divide both sides by cos x
2sinx=√2
sinx=√2/2
x=π/4, 3π/4 (in Q1 and Q2 where sin>0)

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
sin(2theta)=sqrt(2)cos(theta)
---------
sin%282t%29+=+sqrt%282%29%2Acos%28t%29
Use the double angle identity for sin(2t)
2sin%28t%29%2Acos%28t%29+=+sqrt%282%29%2Acos%28t%29
2sin%28t%29%2Acos%28t%29+-+sqrt%282%29%2Acos%28t%29+=+0
cos%28t%29%2A%282sin%28t%29+-+sqrt%282%29%29+=+0
cos(t) = 0
t = pi/2, 3pi/2
---------------
%282sin%28t%29+-+sqrt%282%29%29+=+0
sin%28t%29+=+sqrt%282%29%2F2
t = pi/4, 3pi/4