Question 972252
Find, to the nearest degree, the roots of cos2&#952;-2cos&#952;=0 on the interval 0<&#952;<360
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cos2&#952;-2cos&#952;=0
cos^2(&#952;)-sin^2(&#952;)-2cos(&#952;)
cos^2(&#952;)-1+cos^2(&#952;)-2cos(&#952;)=0
2cos^2(&#952;)-2cos&#952;-1=0
solve for cos&#952; by quadratic formula:
{{{cos(theta) = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=2, b=-2, c=-1
cos&#952;=1.336 (reject, (-1 < cosx <  1)
or
cos&#952;=-0.366
&#952;=111.469&#730;, 248.13&#730;