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 = 2π-1.1278 = 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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