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July 21st, 2013, 12:14 PM   #1
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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!
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July 21st, 2013, 06:40 PM   #2
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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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