Question 585136
What is 2cos^2 &#952; + cos&#952; -1=0 using the interval 0 &#8804; &#952; < 2 &#960;?
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2cos^2 &#952; + cos&#952; -1=0
(2cos&#952;+1)(cos&#952;-1)=0
2cos&#952;+1=0
2cos&#952;=-1
cos&#952;=-1/2
&#952;=2&#960;/3 and 4&#960;/3 (in quadrants II and III where cos<0)
or
cos&#952;-1=0
cos&#952;=1
&#952;=0
ans: 
&#952;=0, 2&#960;/3, and 4&#960;/3