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2Sin(Θ+47°)=1 ΘЄ[0°, 360°)

What I did:
Sin(Θ+47°)=1/2
Sin 1/2 = Θ+47°
30°+360 = Θ+47°
343° = Θ

Ok, so how do i find the other solutions? In this problem its 103°. I think we were suppose to do something with the graphing calculator and something about calculating the zero. I know how to do that, but i don't know how to make it equal 0 so i could put it in Y1? Any other way, what did i do wrong? Is there a prgm that is avilable for the Ti-83+ that can do these problems?

btw:
Θ = theta

whtis the value of 343... figure that out .. and thn u can say that other solutions with be 2 pie ( n)

n wht ever value u figure out in radioans..

That looks like an equation, not an identity. If it were an identity, it would be true for all theta.

1 ΘЄ[0°, 360°)

is supposed to mean

Are you solving 2 Sin(Θ+47°)= 1 ?
If so, Sin(Θ+47°)=0.5
Θ+47° = 30, 150 or 390 degrees
Θ = 103 or 333 degrees if you are limited to theta between 0 and 360 degrees

the ΘЄ[0°, 360°) was suppose to mean its 0° < or = Θ < 360°

I get this now!!, Thanks!!!, i guess i should study now.

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