Question

How do you solve `(tanx + 1)(sinx - 1) = 0`?

Answer 1

`x=-pi/4`, `x=pi/2`

Explanation:

`(tanx+1)(sinx-1)=0`
It means
either `tanx+1=0` or `sinx-1=0`
Therefore we can consider both situations i.e.
`tanx=-1`
`x=tan^-1-1`
`x=-pi/4`
similarly
`sinx-1=0`
`sinx=1`
`x=sin^-1 1`
`x=pi/2`