Find all solutions on the interval (0,2pi):

2-2cos^2=sinx+1

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

2sin^2x - sinx -1 = 0
(2sinx + 1 )(sinx-1) = 0
sinx = -1/2 or sinx = 1

x = 210° or 330° or 90°
or
x = 7π/6 , 11π/6, or π/2

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To solve the equation 2 - 2cos^2(x) = sin(x) + 1 on the interval (0, 2π), we need to find all the values of x that satisfy the equation within this range.

Let's start by simplifying the equation. We can rewrite cos^2(x) as (1 - sin^2(x)) using the Pythagorean identity. Substituting this into the equation, we get:

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

Simplifying further:

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

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

Next, let's rearrange the equation to set it equal to zero:

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

To solve this quadratic equation for sin(x), we can use the quadratic formula:

sin(x) = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 2, b = -1, and c = 1. Plugging in these values, we can solve for sin(x):

sin(x) = (-(-1) ± √((-1)^2 - 4(2)(1))) / (2(2))

sin(x) = (1 ± √(1 - 8)) / 4

sin(x) = (1 ± √(-7)) / 4

Since √(-7) is not a real number, there are no real solutions for sin(x) in this equation. Therefore, there are no solutions within the interval (0, 2π) for the given equation 2 - 2cos^2(x) = sin(x) + 1.