Question 955450
solve equation for solution (0 deg, 360 deg) 4 cos ^2 theta +4 cos theta = 1
4cos^2(x)+4cosx=1
4cos^2(x)+4cosx-1=0
solve for cosx by quadratic formula:
{{{cosx = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=4, b=4, c=-1
ans:
cosx=-1.207 (reject)(-1 ≤ cosx ≤1)
or
cosx=0.207
x≈78.05˚, 281.95˚