How do you solve for x: #cos^2x - sin^2x = sinx#?

#cos^2x-sin^2x=sinx#

#(1-sin^2x)-sin^2x=sinx# #color(white)"ss"# (use Pythagorean Identity)

#1-2sin^2x=sinx#

#0=2sin^2x+sinx-1#

#2sin^2x+sinx-1 = 0#

#(2sinx-1)(sinx+1)=0# #color(white)"ss"# (it may help to substitute
#color(white)"ssssssssssssssssssssssssssssssssss"# #2S^2 + S -1 =0#
#color(white)"ssssssssssssssssssssssssssssssssss"# #(2S-1)( S +1) =0#

So #2sinx -1 =0# OR #sinx+1=0#
and #sinx = 1/2# OR #sinx = -1#

#sinx = 1/2# gives us #x= pi/6 + 2 pi k# or #(5 pi)/6+2 pi k#
#sinx = -1# gives us #x=(3 pi)/2 +2 pi k#
#color(white)"ssssssssssssssssssssssssss"# where #k# is any integer

Solve the equation: f(x) = cos^2 x - sin^2 x - sin x = 0.
Replace cos^2 x by (1 - sin^2 x)
f(x) = 1 - sin^2 x - sin^2 x - sin x = 0. Call t = sin x
Quadratic equation in t: f(t) = -2 t^2 - t + 1 = 0.
Solve this quadratic equation. There are 2 real roots : t1 = -1 and t2 = 1/2.
Solve the basic trig equation: t1 = sin x = -1 --> x = 3Pi/2
Solve t2 = sin x = 1/2 --> x = Pi/6 ; and x = 5Pi/6.
Within period (0. 2Pi), there are 3 answers: Pi/6; 5Pi/6; and 3Pi/2.