precalculus
posted by marie darling on .
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

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. 
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 
Thank you for your help! My numbers were way off haha.