solve the system in interval [o,2π): y=sin^2(x)+cos^2(x) or x+π=cos^-1(y)
To solve the system of equations in the given interval [0, 2π), we will start by simplifying the equations and then solving them one by one.
1. Simplifying the first equation:
y = sin^2(x) + cos^2(x)
Since sin^2(x) + cos^2(x) = 1 for all values of x, we can rewrite the equation as:
y = 1
2. Simplifying the second equation:
x + π = cos^(-1)(y)
cos^(-1)(y) is the inverse cosine function, also known as arccosine. It gives the angle whose cosine is y. Since the range of the arccosine function is [0, π], we need to restrict our solution to the given interval [0, 2π).
3. Substituting y = 1 into the second equation:
x + π = cos^(-1)(1)
x + π = 0
To determine the solution within the given interval, we need to find the value of x that satisfies both equations simultaneously.
From the second equation, we have x + π = 0, which implies x = -π.
Therefore, the solution to the system of equations in the interval [0, 2π) is x = -π since it satisfies both equations:
y = 1 (from the first equation)
x + π = 0 (from the second equation)
Note that since -π is not within the interval [0, 2π), there are no solutions to the system of equations in the given interval.