## anonymous one year ago Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0

1. freckles

$\sin(4x) \neq \sin(2x)+\sin(2x)$

2. ChillOut

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

3. freckles

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

4. freckles

solve both equations

5. 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

6. freckles

on which part are you confused?

7. anonymous

8. freckles

that is only one solution

9. freckles

there are others

10. freckles

can I ask you what you are confused on/

11. freckles

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

12. anonymous

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

13. freckles

where does that equation come from?

14. anonymous

15. 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?

16. freckles

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

17. anonymous

i was confused by your work

18. freckles

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

19. anonymous

2sin u cos u

20. freckles

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

21. freckles

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

22. anonymous

oh ok so the answer would be B?

23. 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?