Question

How do you solve `cos2x+2sin^2x-sinx=0`?

Answer 1

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

Explanation:

As `cos2x=1-2sin^2x` i.e. `cos2x+2sin^2x=1`, we can write

`cos2x+2sin^2x-sinx=0` as

`1-sinx=0`

or `sinx=1`

i.e. `x=2npi+pi/2`, where `n` is an integer.