<PRE><B>Question 881419</B>
Solve the following equation for 0 &amp;#8804; x &amp;#8804; 2&amp;#960;(give answers in exactly in terms of &amp;#960;):
2cos(&amp;#960;/5 - 5&amp;#119909;/2)= &amp;#8730;3
cos(&amp;#960;/5 - 5&amp;#119909;/2)= &amp;#8730;3/2
{{{pi/5-5x/2=pi/6}}} (in quadrant I in which cos&gt;0)
lcd:30
6&amp;#960;-75x=5&amp;#960;
75x=&amp;#960;
x=&amp;#960;/75
..
{{{pi/5-5x/2=11pi/6}}} (in quadrant IV in which cos&gt;0)
lcd:30
6&amp;#960;-75x=55&amp;#960;
75x=-49&amp;#960;
x=-49&amp;#960;/75</PRE>