Answer:
x = arccos(sqrt((1 + sin(x))/2)) + 2npi or x = pi - arccos(sqrt((1 + sin(x))/2)) + 2npi
Explanation:
cos^2(x) - sin^2(x) = sin(x) can be rewritten as cos^2(x) = sin^2(x) + sin(x)
Using the identity sin^2(x) + cos^2(x) = 1, we can simplify the equation as
cos^2(x) = 1 - cos^2(x) + sin(x)
Solving for cos^2(x), we get
cos^2(x) = (1 + sin(x))/2
Therefore, cos(x) = sqrt((1 + sin(x))/2) or cos(x) = -sqrt((1 + sin(x))/2)
To find the possible values of x, we need to find the values of sin(x) that satisfy this equation.
For the domain [0, 2pi), the range of sin(x) is [-1, 1], so we have:
cos(x) = sqrt((1 + sin(x))/2) for -1 <= sin(x) < 1
cos(x) = -sqrt((1 + sin(x))/2) for -1 <= sin(x) < 1
So, the solution is
x = arccos(sqrt((1 + sin(x))/2)) + 2npi or x = pi - arccos(sqrt((1 + sin(x))/2)) + 2npi
where n is an integer.