Question

find solution in the interval [0,2pi): sin(2x)= sqrt3/2

Answer 1

do you have a link?

Answer 2

there's no link?

Answer 3

no

Answer 4

i can rewrite it for you:\[find solution \in the interbal [0,2\pi): \sin(2x)=\frac{ \sqrt{3}}{ 2 }\]

Answer 5

whoops, it supposed to say find in the intervals [0,2pi). Not sure what happened when I wrote the equation lol

Answer 6

If you used the unit circle, what is the value of t that makes sin(t) = sqrt(3)/2 true?

Answer 7

11pi/6

Answer 8

11pi/6 is in Q4 where sine is negative

Answer 9

pi/6?

Answer 10

closer but no

sin(pi/6) = 1/2

Answer 11

we want the sine value to be sqrt(3)/2

Answer 12

pi/3

Answer 13

good, sin(pi/3) = sqrt(3)/2

Answer 14

what other value of t makes sin(t) = sqrt(3)/2 true?

Answer 15

2pi/3

Answer 16

correct

Answer 17

so if

sin(2x) = sqrt(3)/2

then

2x = pi/3 or 2x = 2pi/3

Answer 18

add on "+2pi*n" to capture all coterminal angles

if

sin(2x) = sqrt(3)/2

then

2x = pi/3+2pi*n or 2x = 2pi/3+2pi*n

Answer 19

n is any integer

Answer 20

ok, thank you!

Answer 21

what is the next step

Answer 22

oh theres more? lol..hmm, I don't know. This is the first time I've had a 2x in sine.

Answer 23

divide by 2?

Answer 24

you have 2x = ....

you want x = ...

Answer 25

yes you divide everything by 2

Answer 26

or better yet, multiply everything by 1/2

Answer 27

2x = pi/3+2pi*n

(1/2)*2x = (1/2)*(pi/3)+(1/2)*2pi*n

x = pi/6 + pi*n

Answer 28

pi/6+pi*n and 2pi/6+pi*n?

Answer 29

2x = 2pi/3+2pi*n

(1/2)*2x = (1/2)*(2pi/3)+(1/2)*2pi*n

x = pi/3 + pi*n

Answer 30

ohh! Ok.

Answer 31

if

sin(2x) = sqrt(3)/2

then

2x = pi/3+2pi*n or 2x = 2pi/3+2pi*n

that turns into

x = pi/6 + pi*n or x = pi/3 + pi*n

where n is any integer

Answer 32

you need to now generate particular solutions by plugging in values of n

make sure that x is in the interval [0, 2pi)