Sin(3ø)=1?

Question

solve equation on interval of 0 (less than or equal to) ø < 2π

Answer 1

the idea in this is that
sin(3ø)=1
so 3ø= π/2 + 2π n where n=0,1 ,2 ....
so ø= π/6 +(2π/3) * n
so try n 0,1,2.. until the result of ø>2π
n=0 ø = π/6 (<2π)
n=1 ø= π/6+2π/3 =5π/6 (<2π)
n=2 ø=π/6+4π/3 =9π/6 =3π/2 (<2π)
n=3 ø=π/6+6π/3 =13π/6 (>2π)
so the solutions are {π/6,5π/6 , 9π/6 }

Answer 2

If the sin is 1 then 3ø = pi/2 so ø = pi/6

Answer 3

3Φ = π/2 ± 2πn
Is the general solution: when divided by 3 we get:
Φ = π/6 ± 2πn/3
now you will need to let n = 0,1 and 2 to get all positive solutions up from 0 to 2π
π/6, 5π/6, 7π/6

Answer 4

I have already replied one of your hard problems, you shouldn't be making people do your hws. You will be asked these question on the exam. Don't be a kid, understand and try to solve the problems. We are here to help you when you don't understand. You are fooling yourself, no one else.

Answer 5

theta "ø" is a symbol for "angle".
your question isnt very hard if you have a scientific calculator.

Answer 6

Ans:
Given Sin(3x)=1
Therefore 3x=Sin^(-1)(1)
=Pi/2
Therefore x =Pi/6 which lies b/w 0 and 2pi

Answer 7

theta, the circle witha slash through it, is equal to pi/6. Check: sin(3(pi/6))=sin(3pi/6)=sin(pi/2)=1.

Answer 8

Sin(3ø)=1
3ø = sin^-1(1)
3ø =π/2
ø = π/6,5π/6

Answer 9

3 t= pi/2+2kpi so t= pi/6+2/3*kpi (k=0,1,2)
so t= pi/6,5/6 pi,9/6 pi =3/2pi,