Question

how would you find the intersection between y=cosx and y=sin2x..... thanks :)

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

set them equal
cosx=sin2x
cosx=2sinxcosx
1=2sinx
1/2=sinx
does this help?

Answer 3

careful, you forgot about when cos(x)=0

Answer 4

thanks, but i was just wondering.....how did you get 1 on the left side

Answer 5

It is better to pull the 2sinxcosx to the left and factor

Answer 6

when is sinx 1/2?
when x=pi/6 and when x=5pi/6
so the answer keeps going
so x=pi/6+2npi for n=0,1,2,3,...
and
x=5pi/6+2npi for n=0,1,2,3,...
i divided both sides by cosx to get 1 on that one side

Answer 7

he right to get all of our intersections
we need to bring everything on one side
cosx-2sinxcosx=0
cosx(1-2sinx)=0
cosx=0
1-2sinx=0 => 1/2=sinx (we already solved this one)
cosx=0 when x=pi/2+2npi and x=3pi/2+2npi
n=0,1,2,...

Answer 8

nice call oser

Answer 9

thank you so much oser and myininaya

Answer 10

np

Answer 11

here's the whole kit and kaboodle... (see attached)

Answer 12

thank you so much, that was really nice of you :D

Answer 13

no problem... have a great one! :o)

Answer 14

oser....another question.....how come u multiplied 2sinx with cosx on the right side

Answer 15

one of the trig identities you should know is that sin(2x) = 2sin(x)cos(x). This comes from the addition identity for sines: sin(a+b)=sin(a)cos(b)+sin(b)cos(a). When both a and b equal x, you get sin(2x) = sin(x)cos(x) + sin(x)cos(x) = 2sin(x)cos(x).

Answer 16

oh, wow, i did not know that rule...no wonder i was so confused..thank you so much! u were really helpful

Answer 17

my pleasure. :o)