Find all solutions to `sin^2x+cosx-1=0` :

Use the pythagorean identity to rewrite `sin^2x` as `1-cos^2x`

Thus `1-cos^2x+cosx-1=0` or

`cos^2x-cosx=0`

`cosx(cosx-1)=0`

`=>cosx=0` or `cosx=1`

If `cosx=0` then `x=pi/2+npi` for `ninZZ` (n an integer)

If `cosx=1` then `x=0+2npi` for `ninZZ`

You need to write the equation in terms of cos x only, hence you should use the basic formula of trigonometry `sin^2 x + cos^2 x = 1` .

Substituting the constant term 1 by thesum of squares `sin^2 x + cos^2 x`  yields:

`sin^2 x + cos x - (sin^2 x + cos^2 x)=0`

Opening the brackets yields:

`sin^2 x + cos x - sin^2 x- cos^2 x = 0`

Reducing like terms yields:

`cos x - cos^2 x = 0`

You need to factor out cos x such that:

`cos x*(1 - cos x) = 0 =gt cos x = 0`

`x = +-cos^(-1) (0) + 2npi`

`x = +-pi/2 + 2npi`

`1 - cos x = 0 =gt -cos x = -1`

`cos x = 1 =gt x = +-cos^(-1) (1) + 2npi`

`x = 2npi`

Hence, evaluating all solutions to equation `sin^2 x + cos x - 1= 0`  yields `x = +-pi/2 + 2npi ` and `x = 2npi.`