How do you solve #2sin^2 (x/2)-3sin (x/2) +1=0# ?

#2sin^2 (x/2)-3sin (x/2) +1=0#
for simplicity let #sin(x/2)=A#

now equation is #2A^2-3A+1=0#
#2A^2-2A-A+1=0#
#2A(A-1)-1(A-1)=0#
#(A-1)(2A-1)=0#
either #A=1# or #A=1/2#
either #sin(x/2)=1# or #sin(x/2)=1/2#

if #sin(x/2)=1# then #x/2=sin^-1(1)=pi/2#
hence #x=pi#

if #sin(x/2)=1/2# then #x/2=sin^-1(1/2)=pi/6#
hence #x=pi/3#

Given: #2sin^2(x/2)-3sin(x/2) +1=0#

Please observe that this is exactly like the quadratic:

#2u^2-3u +1=0#

But #u# has been substituted for #sin(x/2)#.

Let's factor the second quadratic:

#(2u-1)(u-1) = 0#

#u = 1/2 and u = 1#

We reverse the substitution:

#sin(x/2) = 1/2 and sin(x/2) = 1#

Use the inverse sine function:

#x/2 = sin^-1(1/2) and x/2 = sin^-1(1)#

The first condition occurs at two values, #pi/6 and (5pi)/6# and the second condition occurs at #pi/2#. All conditions repeat every integer multiple of #2pi#:

#x/2 = pi/6+2npi, x/2 = (5pi)/6+2npi, and x/2 = pi/2+2npi; n in ZZ#

Multiply all three conditions by 2:

#x = pi/3+4npi, x = (5pi)/3+4npi, and x = pi+4npi; n in ZZ#

Call #sin (x/2) = t# and solve this quadratic equation for t
#2t^2 - 3t + 1 = 0#
Since a + b + c = 0, use shortcut. The 2 real roots are:
t = 1 and #t = c/a = 1/2#
a. #t = sin (x/2) = 1# -->
#x/2 = pi/2 + 2kpi# --> #x = pi + 4kpi = (1 + 4k)pi#
b.# t = sin (x/2) = 1/2#
Trig table and unit circle give 2 solutions:
b. #x/2 = pi/6 + 2kpi# --> #x = pi/3 + 4kpi#
c. #x/2 = (5pi)/6 + 2kpi# --> #x = (10pi)/6 = (5pi)/3 + 4kpi#.