Question 966798
Find all angles, 0&#8804;&#952;<360, that satisfy the equation below, to the nearest 10th of a degree.
2cos2&#952;+9=3cos&#952;+8
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{{{2cos(2x)+9=3cosx+8}}}
{{{2(cos^2(x)-sin^2(x))-3cosx+1=0}}}
{{{2(cos^2(x))-(1-cos^2(x))-3cosx+1=0}}}
{{{2cos^2(x)-2+2cos^2(x)-3cosx+1=0}}}
{{{4cos^2(x)-3cosx-1=0}}}
(4cosx+1)(cosx-1)=0
..
4cosx+1=0
cosx=-1/4
x=104.5&#730;, 255.5&#730;
or
cosx+1=0
cosx=-1
x=180&#730;