Question

Find all solutions in the interval [0, 2π).

4 sin2x - 4 sin x + 1 = 0

pi/6, 5pi/6

pi/6, 11pi/6

pi/3, 5pi/3

7pi/6, 11pi/6

Answer 1

hint: let z = sin(x) and make the proper replacements to get 4z^2 - 4z + 1 = 0

Answer 2

let me know if that helps or not

Answer 3

Your equation is a little ambiguous right now. I can't tell if 4sin2x equals 4sin(2x) or 4(sinx)^2.

If it's the former: subtract 1 from both sides, factor a 4 out of the left side, and solve resultant equations.

If it's the latter: treat sinx as any other variable; find the roots (as in a quadratic) and solve resultant equations.

Answer 4

Oh sorry, it's 4sin^2x

Answer 5

Do you know what the answer ended up being?

Answer 6

No sorry, can't remember what it was.

Answer 7

x = pi/6 or 5pi/6 in the interval [0,2pi]