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Mathematics 7 Online
OpenStudy (anonymous):

Find all solutions in the interval [0, 2π). 4 sin2 x - 4 sin x + 1 = 0

OpenStudy (anonymous):

can you factor 4u^2 - 4u + 1 = 0?

OpenStudy (anonymous):

(2x-1)^2

OpenStudy (anonymous):

good. But don't use x. Use a different letter. What do you think 4sin^2(x) - 4sin(x) + 1 is in factor form?

OpenStudy (anonymous):

So.. (2sin x -sin x)^2?

OpenStudy (anonymous):

I will be back in 20.

OpenStudy (anonymous):

I'm back @sourwing !

OpenStudy (anonymous):

Hello? Oh, and it should have been: (2sin x-1)^2. My bad.

OpenStudy (anonymous):

good (2sinx - 1)^2 = 0 what would you do next?

OpenStudy (anonymous):

For these questions what I personally like to do is let sin x=m and the equation will become: 4m^2+4m+1=0 It's a quadratic equation! Just remember to change it back to sin afterwards :)

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