# find all solutions to sin^2x+cosx-1=0

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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.` **

The equation sin^2x+cosx-1=0 has to be solved for x.

sin^2x+cosx-1=0

Replace sin^2x with a square of cos x using the relation sin^2x = 1 - cos^2x

1 - cos^2x + cos x - 1 = 0

-cos^2x + cos x = 0

cos x(-cos x + 1) = 0

cos x = 0 is true for all values of x = 90 + n*360 and 270 + n*360

-cos x + 1 = 0

cos x = 1

This is true for all values of x = n*360 and 180 + n*360

The roots of the equation sin^2x+cosx-1=0 are n*360, 90+n*360, 180 +n*260 and 270 + n*360 degrees.