#2 cos^2 x - sin x - 1 = 0# for
#x in { (3pi)/2+2npi, pi/6+2npi, (5pi)/6+2npi}# where #n in ZZ#
Solve : #2cos^2 x - sin x - 1 = 0# (1)
First, replace #cos^2 x# by #(1 - sin^2 x)#
#2(1 - sin^2 x) - sin x - 1 = 0#.
Call # sin x = t#, we have:
#-2t^2 - t + 1 = 0#.
This is a quadratic equation of the form #at^2+bt+c = 0# that can be solved by shortcut:
#t = (-b +- sqrt(b^2 -4ac))/(2a)#
or factoring to #-(2t-1)(t+1)=0#
One real root is #t_1 = -1# and the other is #t_2 = 1/2#.
Next solve the 2 basic trig functions:
#t_1 = sin x_1 = -1#
#rarr# #x_1 = pi/2 + 2npi# (for #n in ZZ#)
and
#t_2 = sin x_2 = 1/2#
#rarr# #x_2 = pi/6 + 2npi#
or
#rarr# #x_2 = (5pi)/6 + 2npi#
Check with equation (1):
#cos (3pi/2) = 0; sin (3pi/2) = -1#
#x = 3pi/2 rarr 0 + 1 - 1 = 0# (correct)
#cos (pi/6) = (sqrt 3)/2 rarr 2*cos^ 2(pi/6) = 3/2; sin (pi/6) = 1/2#.
#x = pi/6 rarr 3/2 - 1/2 - 1 = 0# (correct)
