How do you solve #Sin(theta)^2 - cos(theta)^2 = sin(theta)#?

Use trig identity #cos 2a = cos^2 a - sin^2 a = 1 - 2sin^2# a, to transform the equation -->
#- (1 - 2sin^2 t) - sin t = 0#
#2sin^2 t - sin t - 1 = 0.#
Solve this quadratic equation for sin t
Since a + b + c = 0, use shortcut. the 2 real roots are: #sin t = 1# and #sin t = c/a = -1/2#.
Trig Table and unit circle -->
a. #sin t = 1# -> #t = pi/2#.
b. #sin t = -1/2# -> 2 solution arcs --> # t = - (pi/6)# and #t = (7pi)/6#
Since the arc #(11pi)/6# is co-terminal to arc #(-pi/6)#, therefor, the answers for #(0, 2pi)# are:
#pi/2, (7pi)/6, and (11pi)/6#

Consider the pythagorean identity #sin^2theta + cos^2theta = 1#. Rearrange this, solving for #-cos^2theta#, and you'll get: #-cos^2theta = sin^2theta - 1#. In order to change all terms to #sin# use this identity:

#sin^2theta + sin^2theta - 1 = sin theta#

Put all terms to one side of the equation.

#2sin^2theta - sin theta - 1 = 0#

Factor:

#2sin^2theta - 2sintheta + sin theta - 1 = 0#

#2sintheta(sin theta - 1) + 1(sin theta - 1) = 0#

#(2sintheta + 1)(sin theta - 1) = 0#

#sintheta = -1/2 and sin theta = 1#

By special angles:

#theta = 210˚ , 330˚ and 90˚#

Hopefully this helps!