English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

I've been working on this problem for like 3 hours, and cannot figure it out. i can do every problem on the page EXCEPT this one. help please lol...

give all positive values of the angle between 0 and 360 degrees to satisfy the equation:
cos2x - sin^2 x/2 + 3/4 = 0

please show me how you did it so I know for future. thanks tons.

2007-11-07 04:23:32 · 4 answers · asked by Fundamenta- list Militant Atheist 5 in Science & Mathematics Mathematics

yes i know the identities.

2007-11-07 04:24:13 · update #1

as soon as i get explained to as to why i can't select a best answer, i'll try to give you my thumbs up lol. thank you guys.

2007-11-07 07:51:26 · update #2

4 answers

cos(2x) = 2cos²(x) -1
sin²(x/2) = (1 - cos(x))/2

2cos²(x) -1 - [(1 - cos(x))/2] + 3/4 = 0
2cos²(x) - 3/2 + (cos(x))/2 + 3/4 = 0
2cos²(x)+ (cos(x))/2 - 3/4 = 0

cos(x) = [-1/2 ± sqrt(1/4 + 6)]/4
cos(x) = 1/2 or cos(x) = -3/4

cos(x) = 1/2 for x=60° or 300°
cos(x) = -3/4 for x ≈ 138.59° or 221.41°

--Live long and prosper--

2007-11-07 04:43:46 · answer #1 · answered by Ron W 7 · 1 0

First, try converting your equation into a polynomial in cos x or sin x:

sin^2 (x/2) = (1 - cos x) / 2 (half-angle sine formula)
This is also an indication that you probably want to convert to cos x rather than sin x, so we convert double-angle cosine formula that way too:
cos 2x = cos^2 x - sin^2 x = 2*cos^2 x - 1

Your equation now becomes a quadratic equation in cos x:
(2*cos^2 x - 1) - (1/2 - 1/2*cos x) + 3/4 = 0
-> 2*cos^2 x + 1/2*cos x - 1 - 1/2 + 3/4 = 0
-> 2*cos^2 x + 1/2*cos x - 3/4 = 0
This can be solved however you prefer to solve a quadratic equation to give cos x = -3/4 or cos x = 1/2. Between 0 and 360 degrees, cos x = 1/2 at 60 degrees and at 300 degrees. The values for cos x = -3/4 aren't quite so pretty, but are approximately 138.6 and 221.4 degrees.

2007-11-07 05:10:53 · answer #2 · answered by devilsadvocate1728 6 · 1 0

cos2x-sin^2(x/2)+3/4=0
2(2cos^2x-1)-(1-cosx)+3/2=0
4cos^2x+cosx-3/2=0
8cos^2x+2cosx-3=0
8cos^2x+6cosx-4cosx-3=0
2cosx(4cosx+3)-1(4cosx+3)=0
(4cosx+3)(2cosx-1)=0
4cosx=-3 or 2cosx=1
2cosx=1
cosx=1/2=cos 60
x=60,300 degrees.
Now if 4cosx=-3
cosx=-3/4=-0.75=-cos41.41
x=180-41.41,180+41.41
= 138.59,221.41 degrees
Answrs;60,138.59,221.41,300 degrees

2007-11-07 04:50:50 · answer #3 · answered by Anonymous · 1 0

just pluggin and chuggin I came up with14.435885
345.564115,
The other people's numbers didn't add up to me.

2007-11-07 06:12:36 · answer #4 · answered by jim m 5 · 0 0

fedest.com, questions and answers