#sin 2x = cos (pi/2 - 2x)#
The equation in question can be transformed to:
#cos (pi/2 - 2x) = cos (x - pi/6)#
Trig unit circle give 2 solutions -->
#(pi/2 - 2x) = +- (x - pi/6)#
a. #pi/2 - 2x = (x - pi/6)#
#3x = pi/2 + pi/6 = 4pi/6 = (2pi)/3#
General answers:
#3x = (2pi)/3 + 2kpi#
#x = (2pi)/9 + (2kpi)/3#
b. #pi/2 - 2x = - x + pi/6# -->
#pi/2 - pi/6 = 2x - x#
#x = pi/2 - pi/6 + 2kpi = (2pi)/6 + 2kpi = (pi/3) + 2kpi#
Check by calculator.
#x = pi/3# --> #sin 2x = sin (2pi/3) = sqrt3/2# -->
#cos (x - pi/6) = cos (pi/6) = sqrt3/2#. Proved
#x = (2pi)/9# --> #sin 2x = sin ((4pi)/9) = sin 80 = 0.98# -->
#cos (x - pi/6) = cos (40 - 30) = cos 10 = 0.98#. Proved.
