Directions: Find all solutions of the equation in the interval (0, 2pi)

sin x/2=1-cosx

recall that

cos (2x) = 1 - 2sin^2 x
or
cos x = 1 - 2sin^2 (x/2)

then yours is ...
sin (x/2) = 1 - (1 - 2sin^2 (x/2))

sin (x/2) = 2sin^2 (x/2)
2sin^2 (x/2) - sin (x/2) = 0
sin x/2(2sin x/2 - 1) = 0

sin (x/2) = 0 or sin(x/2) = 1/2

case 1: sin (x/2) = 0
x/2 = 0 , π, 2π
x = 0, 2π, 4π

case 2: sin (x/2) = 1/2
x/2 = π/6 or 5π/6 , ( 30° or 150°)
x = π/3 , 5π/3

so x = 0, , π/3, 5π/3, 2π for 0 ≤ x ≤ 2π

To find the solutions of the equation sin(x/2) = 1 - cos(x) in the interval (0, 2pi), we'll need to solve the equation for x while considering the given interval.

Let's go step by step:

Step 1: Rewrite the equation using a trigonometric identity.
Since sin(x/2) involves half angle, we can rewrite it in terms of cos(x) using the identity: sin^2(t) = (1 - cos(2t)) / 2.

sin(x/2) = 1 - cos(x)
(sin(x/2))^2 = (1 - cos(x))^2
(1 - cos^2(x/2)) / 2 = (1 - cos(x))^2
(1/2) - (1/2)cos^2(x/2) = 1 - 2cos(x) + cos^2(x)

Step 2: Simplify the equation.
Let's multiply both sides of the equation by 2 to clear the denominator.

1 - cos^2(x/2) = 2 - 4cos(x) + 2cos^2(x)
2cos^2(x) - 4cos(x) + cos^2(x/2) = 1

2cos^2(x) - 4cos(x) + (1 - cos^2(x)) / 2 = 1
2cos^2(x) - 4cos(x) + 1 - cos^2(x) = 2

Combine like terms:
cos^2(x) - 4cos(x) + 1 = 0

Step 3: Solve the quadratic equation.
Now, let's solve the quadratic equation cos^2(x) - 4cos(x) + 1 = 0.

To solve this equation, we can factor it or use the quadratic formula. Factoring may not yield easily factorable solutions in this case, so let's use the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / (2a)

For our equation, a = 1, b = -4, and c = 1.

x = [-(-4) ± sqrt((-4)^2 - 4(1)(1))] / (2*1)
x = (4 ± sqrt(16 - 4)) / 2
x = (4 ± sqrt(12)) / 2
x = (4 ± 2√3) / 2

Simplifying further:
x = 2 ± √3

Therefore, the solutions to the equation sin(x/2) = 1 - cos(x) in the interval (0, 2pi) are x = 2 + √3 and x = 2 - √3.