Question

4 sin^2 x - 4 sin x + 1 = 0 solutions within 2pi and 0

Answer 1

i think it is a perfect square

Answer 2

I know how to graph, but I'm having issues solving algebraically

Answer 3

Yes it is (2sin-1)^2

Answer 4

\[(2\sin(x)-1)^2=0\]

Answer 5

I got pi/6

Answer 6

But I know from graphing that it is pi/6 and 5pi/6

Answer 7

so
\[2sin(x)-1=0\\
2\sin(x)=1\\
\sin(x)=\frac{1}{2}\]

Answer 8

arcsin(1/2) = pi/6

Answer 9

yeah but arcsine only gives you the number on the right side of the unit circle

Answer 10

Then I know I add pi to it to get the other side

Answer 11

no

Answer 12

Reflect over the y?

Answer 13

aka subtract from \(\pi\)

Answer 14

Ok thanks man

Answer 15

Same y though? Isnt cosine x?

Answer 16

yw, always glad to help a ruthless dictator

Answer 17

yeah same y value

Answer 18

And the x's are my solutions

Answer 19

it was sine right

Answer 20

Oh oops

Answer 21

ok maybe that was not a good way to put it
same second coordinate on the unit circle

Answer 22

Makes a lot of sense! Thanks a lot.