Question

i have a question..

Answer 1

about these two..here i have no interval i need to find "all of the solutions"

Answer 2

this is what i did:

Answer 3

\[\sin (3x-\frac{ \pi }{ 3 })=\frac{ 1 }{ 2 }\]
\[3x-\frac{ \pi }{ 3 }=\frac{ \pi }{ 6 }+2kpi\]
\[x=\frac{ \pi }{ 18 }+\frac{ \pi }{ 9 }+\frac{ 2k \pi }{ 3 }\]
\[x=\frac{ \pi }{ 6 }+\frac{ 2k \pi }{ 3 }\]

Answer 4

and
\[3x-\frac{ \pi }{ 3 }=\pi - \frac{ \pi }{ 6 }+2k \pi\]
.
.
.
\[x=\frac{ 7\pi }{ 18 }+\frac{ 2k \pi }{ 3 }\]

Answer 5

so is this it or is there something else that i have to do?

Answer 6

as for the 2nd one i did:
\[\frac{ x }{ 2 }-\frac{ \pi }{ 4 }=\frac{ \pi }{ 3 }+k \pi\]
\[\frac{ x }{ 2 }=\frac{ \pi }{ 3 }+\frac{ \pi }{ 4 }+k \pi\]
\[x=\frac{ 2\pi }{ 3 }\frac{ \pi }{ 2 }+2k \pi\]
\[x=\frac{ 7\pi }{ 6 }+2k \pi\]

Answer 7

Pretty sure that's it.

Answer 8

it is?
okay
i have more xD
sin2x=sinx
2sinx*cosx-sinx=0
2sinx(cosx-1/2)=0
sinx=0 and cosx=1/2

Answer 9

so what do i do there..same thing i need all the solutions
or is that also it?

Answer 10

Alternative:\[\sin(2x) - \sin(x) = 0\]\[\Rightarrow 2\cos(3x/2)\sin(x/2) = 0\]

Answer 11

Your method is fine, though. Just find the values of x.

Answer 12

i use x=arcsina+ 2kpi and x= pi-arcsina+2kpi for sin
x=+- arccosa+2kpi for cos?

Answer 13

Dunno what you're saying but for example if\[\cos(x) = 1/2\]then\[\cos(x) = \pi/3 \pm 2\pi k\]

Answer 14

that's what i wrote xD
so do i just leave it like that or use k as 0,1,2,3 etc
also sinx=0 -> x=k pi what do i do with that?

Answer 15

Great question...do you know the unit circle?

Answer 16

i do :)

Answer 17

That's key.

Answer 18

well sinx could be then pi,2pi,3pi etc...