Question

How do you solve `sin(3x)= -1` with domain between 0 and 2pi?

Answer 1

`sin 3x = -1 = sin ((3pi)/2) -> 3x = (3pi)/2 -> x = pi/2.`

Check: When `x = pi/2 -> 3x - = (3pi)/2 -> sin 3x = -1` OK

Sin 3x= -1 would have us 3x= `(3pi)/2` and also `(3pi)/2 +2pi` and `(3pi)/2+4pi`

x= `pi/2`, `pi/2+(2pi)/3`, `pi/2+(4pi)/3`

x=`pi/2, (7pi)/6, (11pi)/6`

These are the three solution for x in `0<=x<=2pi`

Answer 2

Solve sin 3x = - 1

Ans: `pi/2; (7pi)/6; (11pi)/6` for `(0, 2pi)`

Explanation:

Trig Table of Special Arcs and trig unit circle -->
`sin 3x = - 1 = sin ((3pi)/2)` --> `3x = (3pi)/2 + 2kpi`-->

`x = pi/2 + (2/3)kpi`
- If k = 0 --> `x = pi/2`
- If k = 1 --> `x = pi/2 + (2pi)/3 = (7pi)/6`
- If k = 2 --> `x = pi/2 + (4pi)/3 = (11pi)/6`