#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)