Question

Hi :)
I need to resolve equation f(x)=0 in the intervale (-pi,pi)

f(x)=sin(2x-pi/2)-1/2

I came here but i am not sure if it´s right.
sin(2x)-cos(2x)-3/2=0

Can anyone help, please ? :)

Answer 1

i help you

Answer 2

thanks :)

Answer 3

what grade you are in

Answer 4

plesae

Answer 5

are you still there?

Answer 6

yes

Answer 7

what are are you in

Answer 8

?

Answer 9

-.-

Answer 10

i am in second but i am not sure if it s same as in your country cause i am from Slovakia. I am 16 years old if it helps.

Answer 11

your in second grade

Answer 12

yes

Answer 13

why theying giving you that kind of work

Answer 14

idk :) we have diffrent system (i am in bilingual class)

Answer 15

\[\sin(2x-\frac{\pi}{2})-\frac{1}{2}=0 \\ \sin(2x-\frac{\pi}{2})=\frac{1}{2} \\ \text{ Let } u=2x-\frac{\pi}{2} \\ \text{ so since we had } -\pi<x<\pi \text{ and } x=\frac{u}{2}+\frac{\pi}{4} \\ \text{ Then we need \to solve } \sin(u)=0 \text{ \in the interval } -\pi<\frac{u}{2}+\frac{\pi}{4}<\pi \\ -4\pi<2u+\pi<4\pi \\ -5\pi<2u<3\pi \\ \frac{-5\pi}{2}<u<\frac{3\pi}{2}\]

Answer 16

So you need to solve sin(u)=0 in the interval (-5pi/2,3pi/2)

Answer 17

-.-

Answer 18

This is one way to solve your equation.

Answer 19

i wont even understand that

Answer 20

http://www.tutorpace.com/guest/classroom/4463

Answer 21

thank you :)

Answer 22

hope you seen the type-o above
solve sin(u)=1/2 in the interval (-5pi/2,3pi/2) *

Answer 23

yes thank you really much :) it helped a lot :)