Question 729924: Solve in [0,2pi)
sec(2theta)tan(theta)-sec(2theta)-2tan(theta)+2=0
Please help!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Solve in [0,2pi)
sec(2theta)tan(theta)-sec(2theta)-2tan(theta)+2=0
sec2xtanx-sec2x-2tanx+2=0
sec2x(tanx-1)-2(tanx-1)=0
(tanx-1)(sec2x-2)=0
..
tanx-1=0
tanx=1
x=π/4, 5π/4 (in quadrants I and III where tan>0)
..
(sec2x-2)=0
sec2x=2
cos2x=1/2
2x=π/3, 5π/3 (in quadrants I and IV where cos>0)
x=π/6, 5π/6
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