anonymous
  • anonymous
Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
\[\sin(4x) \neq \sin(2x)+\sin(2x)\]
ChillOut
  • ChillOut
You need to use \(\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\)
freckles
  • freckles
\[\sin(2x)-\sin(4x) =0 \\ \text{ use double angle identity on the } \sin(4x) \\ \sin(2 \cdot [2x])=2 \sin(2x)\cos(2x) \\ \sin(2x)-2 \sin(2x) \cos(2x) =0 \\ \text{ factor out } \sin(2x) \\ \sin(2x)[1-2\cos(2x)]=0 \\ \text{ set both factors equal to 0} \\ \sin(2x)=0 \text{ or } 1-2\cos(2x)=0\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
solve both equations
anonymous
  • anonymous
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
freckles
  • freckles
on which part are you confused?
anonymous
  • anonymous
so the answer is zero?
freckles
  • freckles
that is only one solution
freckles
  • freckles
there are others
freckles
  • freckles
can I ask you what you are confused on/
freckles
  • freckles
my work or solving the equations I asked you to solve?
anonymous
  • anonymous
ok, so I started by making an equation 2x-4x=0
freckles
  • freckles
where does that equation come from?
anonymous
  • anonymous
i made sinx=x
freckles
  • freckles
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?
freckles
  • freckles
sin(x) is only approximately x for values of x near 0 you cannot use this approximation to solve your equation
anonymous
  • anonymous
i was confused by your work
freckles
  • freckles
ok do you know the double angle identities? like for example what does sin(2u)=?
anonymous
  • anonymous
2sin u cos u
freckles
  • freckles
sin(2u)=2sin(u)cos(u) so if u=2x then we have \[\sin(2[2x])=2\sin(2x)\cos(2x)\]
freckles
  • freckles
\[2 \cdot 2 =4 \\ \text{ so } \sin(4x)=\sin(2[2x])=2\sin(2x)\cos(2x) \\ \text{ so I replaced } \sin(4x) \text{ with } 2\sin(2x)\cos(2x)\]
anonymous
  • anonymous
oh ok so the answer would be B?
freckles
  • freckles
\[\sin(2x)-\sin(4x)=0 \\ \sin(2x)-2 \sin(2x) \cos(2x)=0 \\ \text{ next I noticed a common factor in both terms on the left } \\ \color{red}{\sin(2x)}-2 \color{red}{\sin(2x)}\cos(2x)=0\] 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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.