Posted by marie darling on Monday, April 15, 2013 at 8:07pm.
Solve 7 cos(2 theta) = 7 sin^2(theta) + 3 for all solutions 0 less than or equal to theta less than 2pi.
Solve 7 sin(2 w)  4 cos(w) =0 for all solutions 0 le w lt 2pi

precalculus  marie darling, Monday, April 15, 2013 at 8:18pm
le= less than or equal to
lt = less than
I keep getting numbers that are not on the unit circle and then when I try to plug them into my calculator I'm getting really big numbers. Please help. I've tried looking in my book but there are no problems like these.

precalculus  Reiny, Monday, April 15, 2013 at 9:36pm
7 cos(2 theta) = 7 sin^2(theta) + 3
7 cos^2 Ø  7sin^2 Ø = 3
cos^2Ø  sin^2 Ø = 3/7
cos 2Ø = 3/7
so 2Ø is in I or in IV
set your calculator to radians
2Ø = 1.1279 or 2Ø = 2π1.1278 = 5.1553
Ø = .5639 or Ø = 2.578
Now the period of cos 2Ø = π , so by adding multiples of π to any answer will yield a new answer
so Ø = .5639+π = 3.7055
Ø = 5.1553+π = 5.71924
7 sin 2w  4cos w = 0
7(2sinwcosw)  4cosw = 0
cosw(14sinw  4 ) = 0
cosw = 0 or sinw = 2/7
if cosw = 0, then w = π/2 , 3π/2
if cosw = 2/7 , then w = 1.281 or 2π  1.281 = 5.002

precalculus  marie darling, Monday, April 15, 2013 at 10:26pm
Thank you for your help! My numbers were way off haha.
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