Question

How do you solve `2sin^2 theta-sin theta=0`?

Answer 1

`theta = n*pi` or
`theta= pi/6+n*2pi` or
`theta=(5pi)/6 +n*2pi`
`color(white)("XXXXXXXXXXXXXXXX")nin ZZ`

Explanation:

If `2sin^2theta-sin theta = 0`
then
`color(white)("XXX")(sin theta)(2sin theta -1)= 0`

`rarr color(white)("XXX")sin theta=0color(white)("XXXXX")orcolor(white)("XXXXX")2sin theta =1`
`color(white)("XXXX")rarr theta = 0+n*pi color(white)("XXXXXXXX")rarr theta=pi/6+2npi or (5pi)/6+2npi`

Answer 2

I tried to solve it as a normal quadratic equation in `sintheta` instead of `x`:

Explanation:

I would collect `sintheta` and write:

`sintheta[2sintheta-1]=0`

so:

`sintheta=0` when `theta=0,pi,2pi...` or `theta=npi` with `n=0,1,2,3,4....`

and:

`2sintheta-1=0`
`sintheta=1/2` when `theta=pi/6,5/6pi....` and multiples of `2npi` each.