Question

How do you solve `5sinx+2=sinx` for `0<=x<=2pi`?

Answer 1

`x in {pi/6,(11pi)/6}`

Explanation:

We start as if this wasn't a trigonometric equation:
`5sinx+2=sinx` For now we'll 'pretend' that its domain is `D_0=\mathbb{R}` (all `x`s that make sense) and at the end we'll adjust the solutions to its actual domain `D=[0,2pi]`.

`5sinx+2=sinx`
`5sinx-sinx=-2`
`4sinx=-2`
`sinx=-2/4`
`sinx=-1/2`

Now, here we have at least two ways to go, I personally would use the fact that sine is an odd function:

`-sinx=1/2`
`sin(-x)=1/2`

`-x=pi/6+2kpi or -x=pi-pi/6+2kpi, k in mathbbZ`
`x=-pi/6+2kpi or x=-(5pi)/6+2kpi, k in mathbbZ`
(Note: during the last step we multiplied both equations by `(-1)` and yet `+2kpi` stayed `+2kpi` - didn't change to `-2kpi`. The reason for that it that here it doesn't really matter because `k` goes trough the set of all integers - both positive and negative. I'm just used to writing `+somepi`)

Now we can adjust our solution to the domain `D=[0,2pi]`. Concluding:
`x in {(7pi)/6,(11pi)/6}`