Find all the solutions of 2 sinx=1-cosx in the interval from 0° ≤x<360°

square both sides ...

4 sin^2 x = 1 - 2 cosx + cos^2 x
4(1 - cos^2 x) = 1 - 2cosx + cos^2 x
4 - 4cos^2 x = 1 - 2cosx + cos^2 x
5 cos^2 x - 2cosx - 3= 0
(cosx - 1)(5cosx + 3) = 0
cosx = 1 or cosx = -3/5
x = 0 or 360, or x = 126.9° or 233.1°

since I squared, all answers must be checked:

if x=0, LS = 2sin0 = 0
RS = 1 - cos0 = 0, good
if x= 360
LS = 2sin360 = 0
RS = 1 - cos360 = 0, good
if x = 126.9
LS = 2sin126.9 = 1.6
RS = 1 - cos126.9 = 1-(-3/5) = 1.6 , good
if x = 233.1°
LS = 2sin233.1 = -1.6
RS = 1 - cos233.1 = 1.6 , no good

x = 0, 126.9, and 360

except 0°≤x<360°

To find all the solutions of the equation 2sin(x) = 1 - cos(x) in the interval from 0° ≤ x < 360°, we need to simplify the equation and then solve for x.

Let's start by simplifying the equation. We'll convert sin(x) and cos(x) into their respective double-angle formulas:

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

Now, let's simplify further:

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

Rearranging the terms:

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

Now, we can factor out sin(x) from the equation:

sin(x)(2sin(x) - 2) = 0

We have two possible conditions for the equation to be true: sin(x) = 0 or 2sin(x) - 2 = 0.

Condition 1: sin(x) = 0
In this case, we know that sin(x) = 0 when x is an integer multiple of 180°, or x = 0°, 180°, and 360°.

Condition 2: 2sin(x) - 2 = 0
We'll solve this equation separately:

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

We know that sin(x) = 1 when x = 90°.

So, the possible solutions in the interval 0° ≤ x < 360° are x = 0°, 90°, 180°, and 360°.

To summarize, the solutions to the equation 2sin(x) = 1 - cos(x) in the interval from 0° ≤ x < 360° are x = 0°, 90°, 180°, and 360°.