Question

How do you solve `cos^2x = 1 + sinx` on the interval [0,2pi]?

Answer 1

`x= 0, pi, 2pi and (3pi)/2`

Explanation:

`cos^2 x = 1 - sin^2 x`, hence the given equation would become `1-sin^2 x = 1+ sin x`

Or, `" "sin^2 x +sin x=0`

Or, `" "sin x (sinx +1)=0`

This give `sin x=0` and `sin x=-1`

`x= 0, pi, 2pi and (3pi)/2`