Question

How do you solve `sinx-2sinxcosx=0`?

Answer 1

`x=npi` or `x=2npi+-pi/3`, where `n` is an integer

Explanation:

`sinx-2sinxcosx=0`

`hArrsinx(1-2cosx)=0`

Hence either `sinx=0` i.e. `x=npi`, where `n` is an integer

or `1-2cosx=0` i.e. `cosx=1/2` and `x=2npi+-pi/3`, where `n` is an integer