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    <title>Recent Posts in 'A maths! Trigonometric questions!' | sgForums.com</title>
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      <title>A maths! Trigonometric questions! replied by eagle @ Sun, 07 Sep 2008 19:04:30 +0800</title>
      <description>&lt;p&gt;cot 2x = cos 2x * cosec 2x&lt;/p&gt;
&lt;p&gt;hence, 8 cosec 2x cos 2x cosec 2x = 3&lt;/p&gt;
&lt;p&gt;8 cos 2x = 3 (sin 2x )^2&lt;br /&gt;
8 cos 2x = 3 - 3 (cos 2x )^2&lt;br /&gt;
3 (cos 2x )^2 + 8 cos 2x - 3 = 0&lt;br /&gt;
(3 cos 2x -1) (cos 2x + 3) = 0&lt;br /&gt;
cos 2x = 1/3&amp;nbsp; or&amp;nbsp; cos 2x = -3 (reject because cos is
between -1 and 1)&lt;/p&gt;
&lt;p&gt;0 deg &amp;lt; x &amp;lt; 180 deg&lt;br /&gt;
0 deg &amp;lt; 2x &amp;lt; 360 deg&lt;/p&gt;
&lt;p&gt;so 2x = 70.53, 298.47&lt;br /&gt;
x = 35.3 deg, 144.7 deg&lt;/p&gt;
&lt;p&gt;Regards,&lt;br /&gt;
Eagle (&lt;a href="http://www.freewebs.com/strategictuition" rel=
"nofollow"&gt;Strategic Tuition&lt;/a&gt;)&lt;/p&gt;</description>
      <pubDate>Sun, 07 Sep 2008 19:04:30 +0800</pubDate>
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      <author>eagle</author>
      <link>http://weewee.sgforums.com/forums/2297/topics/329836</link>
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      <title>A maths! Trigonometric questions! replied by bonkysleuth @ Sun, 07 Sep 2008 18:39:13 +0800</title>
      <description>&lt;p&gt;Need help with this question. Here goes:&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Find the values of x, where 0 deg &amp;lt; x &amp;lt; 180 deg, such
that&lt;/p&gt;
&lt;p&gt;8 cosec 2x cot 2x = 3&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Thanks guys!&lt;/p&gt;</description>
      <pubDate>Sun, 07 Sep 2008 18:39:13 +0800</pubDate>
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      <author>bonkysleuth</author>
      <link>http://weewee.sgforums.com/forums/2297/topics/329836</link>
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