Use trig identity:
sin 2x = 2sin x.cos x.
In this case:
#sin^2 x = 4sin^2 (x/2).cos^2 (x/2)#
Rewrite the equation:
#4sin^2 (x/2).cos^2 (x/2) - cos^2 (x/2) = 0#
#cos^2 (x/2)(4sin^2 (x/2) - 1) = 0#
Either factor must be zero.
a. #cos^2 (x/2) = 0# --> #cos (x/2) = 0#
Unit circle gives 2 solutions:
#x/2 = pi/2 + 2kpi# --> #x = pi + 4kpi#, and
#x/2 = (3pi)/2 + 2kpi#--> #x = 3pi + 4kpi#
b. #4sin^2 (x/2) = 1# --> #sin^2 (x/2) = 1/4#
#sin (x/2) = +- 1/2#
Trig table and unit circle give 2 solutions:
a. #sin x = 1/2# --> #x/2 = pi/6 + 2kpi# --> #x = pi/3 + 4kpi#
and #x/2 = (5pi)6 + 2kpi# --> #x = (5pi)/3 + 4kpi#
b. #sin x = - 1/2# --> #x/2 = (7pi)/6 + 2kpi# -->
#x = (7pi)/3 = (pi/3) + 4kpi# , and
#x/2 = (11pi)/6 + 2kpi# --> #x = (11pi)/3 = (5pi)/3 + 4kpi#
