Question

Question #19e54

Answer 1

`2sin^2(theta) = 1, theta in [0,2pi] <=> theta in {pi/4, (3pi)/4, (5pi)/4, (7pi)/4}`

Explanation:

`2sin^2(theta) = 1`

`=> sin^2(theta) = 1/2`

`=> sin(theta) = +-sqrt(1/2) = +-sqrt(2)/2`

If we look at our unit circle, we find that `sin(theta) = sqrt(2)/2` for `theta in {pi/4, (3pi)/4}` and `sin(theta) = -sqrt(2)/2` for `theta in {(5pi)/4,(7pi)/4}`. Thus, taking the union of the sets, we find the result to be

`2sin^2(theta) = 1, theta in [0,2pi] <=> theta in {pi/4, (3pi)/4, (5pi)/4, (7pi)/4}`