Question

How do you solve `cos2x-sinx=1` in the interval [0,360]?

Answer 1

The answer is `S={0,180,210,330}º`

Explanation:

We use

`cos2x=1-2sin^2x`

The equation becomes

`1-2sin^2x-sinx=1`

`2sin^2x+sinx=0`

`sinx(2sinx+1)=0`

So, `sinx=0` or `sinx=-1/2`

`x=0, 180`º and `x=210, 330`º