Solve sin theta=-1 for all real values of theta.
Is it theta=3pi/2 +2pi n for n=0+ infinity?
Yes, but n can be any integer from -infinity to infinity
thank you
Yes, you are correct. The solution to the equation sin(theta) = -1 for all real values of theta is indeed theta = (3π/2) + 2πn, where n is an integer that can take any non-negative value (0, 1, 2, 3, ...).
To explain how to arrive at this solution, we know that the sine function has a value of -1 at the angle (3π/2) plus even multiples of π. In other words, the angle (3π/2) + 2πn gives us -1 as the sine function's output for any integer value of n.
To understand why this is the case, let's visualize the unit circle. The point on the unit circle corresponding to an angle of (3π/2) is located at the coordinates (0, -1), which is the point directly below the center of the circle. The sine function gives the y-coordinate of a point on the unit circle, so sin((3π/2) + 2πn) = -1 for any integer value of n.
Hence, the solution to sin(theta) = -1 for all real values of theta is given by theta = (3π/2) + 2πn, where n is an integer.