Question

How do you solve `sinxcosx+cosx=0`?

Answer 1

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

Explanation:

`sinxcosx+cosx=0`

`hArrcosx(sinx+1)=0`

i.e. either `cosx=0` and `x=(npi+-pi/2)`

or `sinx=-1` and `x=2npi-pi/2`

where `n` is an integer

As the two are contained in `x=(npi+-pi/2)`, where `n` is an integer, this is the solution.