Question

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

Answer 1

`x=pi` and `x=pi/3`

Explanation:

`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`

Answer 2

Substitute `u = sin(x/2)`.
Solve the quadratic.
Reverse the substitution
Use the inverse sine function.
Add the periodic nature to the answer.

Explanation:

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`

Answer 3

`x = (1 +4k)pi`
`x = pi/3 + 4kpi`
`x = (5pi)/3 + 4kpi`

Explanation:

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`.