Question

Solve for the interval 2sin2x-1=0 interval[0,2pi)

Answer 1

where are you stuck?

Answer 2

I found out the answer and it is {pi/12, 5pi/12, 13pi/12, and 17pi/12} I do not understand how to get the 13pi/12 and 17pi/12 !

Answer 3

ok, so you get to
sin2x=1/2
2x=????

Answer 4

that equals 1

Answer 5

no...
2x=arcsin(1/2)

what is the inverse sine of 1/2 ?

Answer 6

i.e. at what angle is the sine 1/2 ?

Answer 7

30 degrees, or \[\pi/6, 5\pi/6\]

Answer 8

great, so we have\[2x=\{\frac\pi 6,\frac{5\pi} 6\}\] but any time we add \(2\pi\) we will also have the same point on the unit circle, so we can write\[2x=\{\frac\pi 6+2\pi n,\frac{5\pi} 6+2\pi n\};n \in\mathbb N\]

Answer 9

so far so good?

Answer 10

yep!

Answer 11

now we solve for x\[x=\{\frac\pi{12}+\pi n,\frac{5\pi}{12}+\pi n\}; n\in\mathbb N\]do you see how the other answers come up now?

Answer 12

so I just pick any number, n, and plug it into that equation. I just tried to use n=2 and got \[25\pi/12\]

Answer 13

ah, but you have missed the fact that when we divided by 2, the +2pi*n became +pi*n

plug in n=0 and n=1 now and see what you get

Answer 14

if you use n=2 that takes us out of the range of 0 to 2pi, so that is invalid here

Answer 15

you can't just use ANY n to get all answers.... I think I misunderstood the problem you were having

Answer 16

n=0, ='s \[\pi/12\] and n=1, ='s \[13\pi/12\]

Answer 17

do I not try to make n=3 or n=4 because the interval only includes 2 pi?

Answer 18

yes, you have to check how many values of n (n=0,1,2,3) can fit until you leave the range 0<x<2pi

Answer 19

in this case only n=0 and n=1 give angles in the range specified

Answer 20

and how do I tell is something is more than 2pi?

Answer 21

well with n=0 your solutions are pi/12 and 5pi/2
with n=1 they are 13pi/12 and 17pi/12
with n=2 they are 25pi/12 and 29pi/12, both of which are greater than 2pi (since 25/12>2)

Answer 22

great thank you sooooo much this was very helpful :)

Answer 23

Glad I could help :)
See ya around!