The problem is:
Solve each of the following equations over the interval [0,2Pi).
2sin^2 x - 1 = 0
The answers are:
Pi/4, 3Pi/4, 5Pi/4, 7Pi/4
I know how to get the first two answers, but I'm not sure how you get the last two answers, since sin would be negative in quadrants 3 and 4.
Much help appreciated. Thanks.
Well, remember that a square root has both +- roots.
Wouldn't the last two answers be negative then?
To solve the equation 2sin^2(x) - 1 = 0 over the interval [0, 2Pi), we can follow these steps:
1. Start by isolating sin^2(x) on one side of the equation:
2sin^2(x) - 1 = 0
2sin^2(x) = 1
sin^2(x) = 1/2
2. Take the square root of both sides to solve for sin(x):
sin(x) = ±√(1/2)
3. Now, we need to find the angles within the interval [0, 2Pi) where the sine function takes on the value of ±√(1/2).
For sin(x) = √(1/2), we can determine the angles by taking the inverse sine (sin^-1) of √(1/2):
x = sin^-1(√(1/2))
x = π/4 or x = 3π/4
However, we need to be mindful of the given interval [0, 2Pi), so we only consider the values that fall within this range. Hence, the solutions for sin(x) = √(1/2) within [0, 2Pi) are π/4 and 3π/4.
For sin(x) = -√(1/2), we can determine the angles by taking the inverse sine (sin^-1) of -√(1/2):
x = sin^-1(-√(1/2))
x = 5π/4 or x = 7π/4
Again, we consider the values within the interval [0, 2Pi), so the solutions for sin(x) = -√(1/2) within [0, 2Pi) are 5π/4 and 7π/4.
Hence, the solutions to the equation 2sin^2(x) - 1 = 0 over the interval [0, 2Pi) are:
x = π/4, 3π/4, 5π/4, 7π/4.