Warning: Illegal string offset 'html' in /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php on line 909

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 114

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 127

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 136

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 137

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 141
2003 Paper Two - HSN forum

Jump to content


2003 Paper Two


2 replies to this topic

#1 SomethingTypical

    Site Swot

  • Members
  • PipPipPipPip
  • 111 posts
  • Location:Caithness
  • Gender:Male

Posted 18 May 2005 - 05:54 PM

I've been trying to work along with the solutions on mathsrevision.com but I can't follow it.

Some help please? sad.gif

#2 eck231

    Showing Improvement

  • Members
  • PipPip
  • 20 posts

Posted 18 May 2005 - 06:50 PM

hey

the cos2x means u can use the formula sheet, choosing the version that'l match the other trig terms

3cos2x + 10cosx - 1 = 0

3(2cos²x-1) + 10cosx - 1 = 0

6cos²x - 3 + 10cosx - 1 = 0

6cos²x + 10cosx - 4 = 0

You can take out a common factor here to make it easier to factorise, so it becomes

3cos²x + 5cosx - 2 = 0
(3cosx - 1) (cosx + 2) = 0

cosx = 1/3 or -2

At the start of the quesition it states cos x must lie between -1 and 1 so use

cosx = 1/3

x = cos-¹ (1/3)
x = 1.2309...

So x = 1.23

Hope this helps! biggrin.gif


#3 eck231

    Showing Improvement

  • Members
  • PipPip
  • 20 posts

Posted 19 May 2005 - 03:51 PM

sorry posted by mistake!





1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users