Start from the given #2 sin^2 x - cos x = 1#
note that, #sin^2 x+cos^2 x=1# and #sin^2 x=1-cos^2 x#
and #2 sin^2 x - cos x = 1#
is also #2*(1-cos^2 x)-cos x=1#
and
#2-2cos^2 x-cos x=1#
#2 cos^2 x + cos x-2+1=0#
#2 cos^2 x + cos x-1=0#
use #color(red)("Quadratic Equation")#
#cos x=(-b+-sqrt(b^2-4ac))/(2a)#
#2 cos^2 x + cos x-1=0#
from #2 cos^2 x + 1*cos x-1=0#
Let #a=2#, #b=1#, #c=-1#
#cos x=(-b+-sqrt(b^2-4ac))/(2a)#
#cos x=(-1+-sqrt(1^2-4*(2)*(-1)))/(2*2)#
#cos x=(-1+-3)/4#
#cos x_1=1/2# and #cos x_2=-1#
#x_1=60^@, 300^@#
#x_2=180^@#
God bless America ....
