Question

trigonometric equations

Answer 1

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

4 sin2x - 4 sin x + 1 = 0

Answer 2

can some one help me even begin to set this up?

Answer 3

http://www.wolframalpha.com/input/?i=4+sin2x+-+4+sin+x+%2B+1+%3D+0

Answer 4

\[4 \sin ^2x-4 \sin x+1=0\]
\[\left( a-b \right)^2=a^2-2*a*b+b^2\]

Answer 5

what is that?

Answer 6

\[a=2 \sin x,b=1\]
or you can make factors
write -4 sin x=-2 sin x-2 sin x

Answer 7

wouldnt it be a=4sinx?

Answer 8

\[\left( 2 \sin x \right)^2=4\sin ^2x\]

Answer 9

so its like (2sinx-1)^2=(2sinx)^2-2*2sinx*1+(2sinx)^2

Answer 10

correct

Answer 11

okay and then what

Answer 12

wait last term should be \[1^2 \]

Answer 13

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

Answer 14

okay so (2sinx-1)^2=(2sinx)^2-2*2sinx*1+1^2
instead. now what?

Answer 15

i have written above.
\[\sin x=\frac{ 1 }{ 2 }=\sin 30,\sin \left( 180-30 \right)\]

Answer 16

x=?

Answer 17

woah, where in the world did you get those big numbers from a few 2's

Answer 18

x=30,150 in degrees.

Answer 19

... where did you get 30 from?

Answer 20

\[\sin x=\frac{ 1 }{ 2 }=\sin \frac{ \pi }{ 6 },\sin \left( \pi-\frac{ \pi }{ 6 } \right)=\sin \frac{ \pi } { 6 },\sin \frac{ 5\pi }{ 6 }\]
\[x=\frac{ \pi }{ 6 },\frac{ 5\pi }{ 6 }\]

Answer 21

sin30 or sin pi/6=1/2

Answer 22

\[\sin \left( 180-\theta \right)=\sin \theta \]

Answer 23

im so screwed in this class.... .-. i have no idea what that means

Answer 24

\[\sin \left( \pi-\theta \right)=\sin \theta \]

Answer 25

i am leaving now.

Answer 26

okay.... thanks?

Answer 27

yw