Let t = sin x - cos x ==--> #(sin x - cos x)^2 = t^2 = #
#= sin^2 x + cos^2 x - 2sin x.cos x =# -->
#2sin x.cos x = 1 - t^2#
Back to equation (1), we get:
#2(1 - t^2) - 2sqrt3(sin x - cos x) - 3 = 0#
#2 - 2t^2 - 2sqrt3t - 3 = 0#. By changing side, we get a quadratic equation in t:
#2t^2 + 2sqr3t + 1 = 0# (2)
#D = d^2 = b^2 - 4ac = 12 - 8 = 4# --> #d = +- 2#
The 2 real roots are:# t = -sqrt3/2 +- 1/2 = (-sqrt3 +-1)/2#
a. #sin x - cos x = t = (-sqrt3 + 1)/2 = -0.732/2 = -0.37#
Reminder of trig identity: #sin x - cos x = -sqrt2cos(x + 45)#. Therefor,
#- sqrt2cos (x + 45) = - 0.37# -->#cos (x + 45) = 0.37/1.414 = 0.26#
#x + 45 = +- 74.83# (#cos +- 74.83 = 0.26#)
#x + 45 = 74.83# --># x = 74.83 - 45 = 29.83# deg
#x + 45 = -74.83# --> #x = -119.83# deg
#b. -sqrt2(x + 45) = (-sqrt3 - 1)/2 = - 2.732/2= #
#(x + 45) = (2.732/4.828) = 0.57#
#(x + 45) = +- 55.54# deg
#x + 45 = 55.54 # -> #x = 55.54 - 45 = 10.54#
#x + 45 = -55.54# -->#x = -55.54 - 45 = -100.54# deg
