Report, edit, etc...Posted by Gigins on 2006-12-17 at 09:12:31
Solve this please. 
2cos[sup]2[/sup](x-3Pi/2)+5sin(x-Pi/2)=4
and this one.
sin(90°-x)-2cos[sup]2[/sup]x=-1
2cos[sup]2[/sup](x-3Pi/2)+5sin(x-Pi/2)=4
and this one.
sin(90°-x)-2cos[sup]2[/sup]x=-1
Report, edit, etc...Posted by Yenku on 2006-12-17 at 09:24:12
Whats this for? You should probably do this yourself, especially if you hope to do well.
Give me a good reason and I might before tomorrow.
Give me a good reason and I might before tomorrow.
Report, edit, etc...Posted by Gigins on 2006-12-17 at 10:26:36
I have no idea how to solve them. I have a hole in my notes right when we went through this theme because I wasn't at school. Also I don't have any book about this stuff.
I need to solve this so I can get a grade in math at all. Please help.
I need to solve this so I can get a grade in math at all. Please help.
Report, edit, etc...Posted by JaFF on 2006-12-17 at 10:31:34
I wish I could, but we didn't have this theme yet.
DEAD, you have a rule that you must have all tests written to get a mark on the subject? Or your math is in such an ars that ony this can save you?
DEAD, you have a rule that you must have all tests written to get a mark on the subject? Or your math is in such an ars that ony this can save you?
Report, edit, etc...Posted by Gigins on 2006-12-17 at 10:36:58
I don't exactly need all the tests written, but without this I'm screwed. 
Report, edit, etc...Posted by JaFF on 2006-12-17 at 10:45:00
You can try and google it, or go to the library.
Good luck.
Good luck.
Report, edit, etc...Posted by DT_Battlekruser on 2006-12-17 at 13:09:24
QUOTE(Gigins @ Dec 17 2006, 06:12 AM)
Solve this please.
2cos[sup]2[/sup](x-3Pi/2)+5sin(x-Pi/2)=4
and this one.
sin(90°-x)-2cos[sup]2[/sup]x=-1
[right][snapback]604772[/snapback][/right]
2cos[sup]2[/sup](x-3Pi/2)+5sin(x-Pi/2)=4
and this one.
sin(90°-x)-2cos[sup]2[/sup]x=-1
[right][snapback]604772[/snapback][/right]
2cos[sup]2[/sup](x - 3π/2) + 5sin(x - π/2) = 4
2cos[sup]2[/sup](x - 3π/2) + 5sin(-(π/2 - x)) = 4
2cos[sup]2[/sup](x - 3π/2) - 5sin(π/2 - x) = 4
2cos[sup]2[/sup](x - 3π/2) - 5cosx = 4
1 + cos(2x - 3π) - 5cosx = 4
sin(π/2 - x) - 2cos[sup]2[/sup]x = -1
cosx - 2cos[sup]2[/sup]x = -1
2cos[sup]2[/sup]x - cosx - 1 = 0
(2cosx + 1)(cosx - 1) = 0
2cosx = -1, cosx = 1
cosx = -1/2, cosx = 1
x = {0 + 2nπ, 2π/3 + 2nπ, 4π/3 + 2nπ}, n є I
Hmm, I'm probably just being stupid on the first one, but #2 was pretty easy.
Report, edit, etc...Posted by Gigins on 2006-12-17 at 13:50:38
Hmm thanks a lot DTBK. I just wonder why is 5sin(π/2 - x) the same as 5cosx. Reduction, right?
Report, edit, etc...Posted by HolySin on 2006-12-17 at 15:34:56
It's because sin(π/2) = cos(0).
Report, edit, etc...Posted by Gigins on 2006-12-17 at 15:59:31
Oh cool. Hmm also I don't get this.
2cos2x - cosx - 1 = 0
(2cosx + 1)(cosx - 1) = 0
2cos2x - cosx - 1 = 0
(2cosx + 1)(cosx - 1) = 0
Report, edit, etc...Posted by DT_Battlekruser on 2006-12-17 at 16:29:31
Factoring?
Report, edit, etc...Posted by Yenku on 2006-12-17 at 20:58:37
I think you made an error in the first one. Go a step back, whats cos(x-3pi/2) = ?
The first one is just lame.
The second one looks good DTBK.
The first one is just lame.
The second one looks good DTBK.
Report, edit, etc...Posted by Ultramilkman on 2006-12-17 at 22:46:48
Yahoo answers. Trust me it works.