Question

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

Answer 1

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.

Answer 2

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 = 2π-1.1278 = 5.1553
<b>Ø = .5639 or Ø = 2.578</b>
Now the period of cos 2Ø = π , so by adding multiples of π to any answer will yield a new answer

so <b>Ø = .5639+π = 3.7055
Ø = 5.1553+π = 5.71924</b>


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 = <b>π/2 , 3π/2</b>
if cosw = 2/7 , then w = <b>1.281</b> or 2π - 1.281 = <b>5.002</b>

Answer 3

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