Solve sin^2(2x-pi/3)=1/4 for all solutions in the interval [0,2pi)

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Solve sin^2(2x-pi/3)=1/4 for all solutions in the interval [0,2pi)

Oct 19th, 2015

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sin (2x-pi/3)= 1/2 or sin (2x-pi/3)= -1/2

for sin (2x-pi/3)= 1/2, the solutions satisfies

2x-pi/3 = pi/6, 2x-pi/3 = 5pi/6, 2x-pi/3 = 13/6 pi or 2x-pi/3 = 17/6 pi 

the solututions are 

x = pi/4, x = 7pi/12, x = 5pi/4, or 19pi/12

for sin (2x-pi/3)= -1/2, the solutions satisfies

 2x-pi/3 = 7pi/6, 2x-pi/3 = 11pi/6, 2x-pi/3 = -1/6 pi or 2x-pi/3 = 19pi/6 pi 

the solutions are 

x = 3/4pi, x = 13/12 pi, x = pi/12, x=7/4pi


Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 19th, 2015

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