Question

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

4 sin2x - 4 sin x + 1 = 0

Answer 1

pi/6, 5pi/6
pi/6, 11pi/6
pi/3, 5pi/3
7pi/6, 11pi/6

Answer 2

Is that 4sin(2x), or 4sin^2(x). If it is the sin squared, can you not let sin(x) = u, solve it like a quadratic, back sub in sin(x) for u, find the angles that correspond to those values, and then check that they fall within your given interval?

Answer 3

4 sin^2 x - 4 sin x +1 = 0

Answer 4

Right, yeah, so can you follow my suggestion or would you like to see some of it worked?

Answer 5

i just need the answer like soon

Answer 6

Hrm, we better learn how to solve it then! Okay, so if we let sin(x) = u, what will your equation look like then?

Answer 7

idk

Answer 8

I'm thinking 4u^2 - 4u + 1 = 0. How does that look?

Answer 9

oh, good

Answer 10

Okay, now we need to solve this quadratic. Let me know what you come up with.