Question

How do you solve `sinxtan^2x=sinx`?

Answer 1

`x=npi` or `x=npi+-pi/4` where `n` is an integer

Explanation:

As `sinxtan^2x=sinx`

`sinxtan^2x-sinx=0`

or `sinx(tan^2x-1)=0`

i.e. either `sinx=0=sin0` i.e. `x=npi`

or `tan^2x=1=tan^2(pi/4)`

i.e. `tanx=+-tan(pi/4)`or `x=npi+-pi/4`

where `n` is an integer