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July 21st, 2013, 01:14 PM  #1 
Newbie Joined: Jul 2013 Posts: 5 Thanks: 0  Trigonometry Homework problem.
So I started my homework, and I would like opinions if I did it right. Given: Solve the following trigonometric equation finding all solutions for x in the interval [0, 2pi] 2cos(2x)csc^2(x)=2cos(2x) So I moved the 2cos(2x) to the other side, 2cos(2x)csc^2(x)2cos(2x)=0 So then I simplify; 2cos(2x)(csc^2(x)1)=0 Then I solved for the Os cos2x=0 csc^2(x)=0 cos in 0 will be pi/2 , 3pi/2 csc= +1 (then I convert to the reciprocal sin=+ 1) then pi/2, 3pi/2 So my answers will be x=pi/4+pik K=0=pi/4 k=1=pi k=2=9pi/4 (this out of range) Please guys or gals check my problem and tell me if I did it right! 
July 21st, 2013, 07:40 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,422 Thanks: 1462 
You made a reasonable start, but then didn't post all the steps of your work, so I don't know precisely how you made certain errors. There are six solutions in the specified interval, but ? is not a solution.


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