Question

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

sin 2x - sin 4x = 0

Answer 1

\[\sin(4x) \neq \sin(2x)+\sin(2x)\]

Answer 2

You need to use \(\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\)

Answer 3

\[\sin(2x)-\sin(4x) =0 \\ \text{ use double angle identity on the } \sin(4x) \\ \sin(2 \cdot [2x])=2 \sin(2x)\cos(2x) \\ \sin(2x)-2 \sin(2x) \cos(2x) =0 \\ \text{ factor out } \sin(2x) \\ \sin(2x)[1-2\cos(2x)]=0 \\ \text{ set both factors equal to 0} \\ \sin(2x)=0 \text{ or } 1-2\cos(2x)=0\]

Answer 4

solve both equations

Answer 5

wait, I'm so confused. here are the options i have
A) π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6
B) 0, π/6, π/2, 5π/6, π, 7π/6, 3π/2, 11π/6
C) 0, 2π/3, 4π/3
D) 0, π/3, 2π/3, π, 4π/3, 5π/3

Answer 6

on which part are you confused?

Answer 7

so the answer is zero?

Answer 8

that is only one solution

Answer 9

there are others

Answer 10

can I ask you what you are confused on/

Answer 11

my work
or solving the equations I asked you to solve?

Answer 12

ok, so I started by making an equation
2x-4x=0

Answer 13

where does that equation come from?

Answer 14

i made sinx=x

Answer 15

is there anyway you can tell me what you are confused on? was it my work or me asking you to solve those two equations I asked you to solve?

Answer 16

sin(x) is only approximately x for values of x near 0
you cannot use this approximation to solve your equation

Answer 17

i was confused by your work

Answer 18

ok do you know the double angle identities?
like for example what does sin(2u)=?

Answer 19

2sin u cos u

Answer 20

sin(2u)=2sin(u)cos(u)
so if u=2x then we have
\[\sin(2[2x])=2\sin(2x)\cos(2x)\]

Answer 21

\[2 \cdot 2 =4 \\ \text{ so } \sin(4x)=\sin(2[2x])=2\sin(2x)\cos(2x) \\ \text{ so I replaced } \sin(4x) \text{ with } 2\sin(2x)\cos(2x)\]

Answer 22

oh ok
so the answer would be B?

Answer 23

\[\sin(2x)-\sin(4x)=0 \\ \sin(2x)-2 \sin(2x) \cos(2x)=0 \\ \text{ next I noticed a common factor in both terms on the left } \\ \color{red}{\sin(2x)}-2 \color{red}{\sin(2x)}\cos(2x)=0\]
if you want to put a 1 next to sin(2x) you can
you can do this since 1*sin(2x) is sin(2x)
\[1 \cdot \color{red}{\sin(2x)}-2 \color{red}{\sin(2x)}\cos(2x)=0\]
now factor out the thing in red
\[\color{red}{\sin(2x)}[1-2 \cos(2x)]=0 \\ \text{ so we have to solve both of the following equations } \\ \sin(2x)=0 \text{ or } 1-2\cos(2x)=0\]

if you want to replace 2x with u
and solve for u first then you can come back later and solve for x but replacing u back with 2x



\[\sin(u)=0 \text{ or } 1-2 \cos(u)=0 \\ \sin(u) =0 \text{ or } \cos(u)=\frac{1}{2}\]
on the unit circle when do you have the y-coordinates are 0?
on the unit circle when do you have the x-coordinates are 1/2?