Question

How do you solve `sin^2A=cosA-1`?

Answer 1

`A=k * 2pi, k in ZZ`

Explanation:

Based on the Pythagorean Theorem we know
`color(white)("XXX")sin^2(A)+cos^2(A)=1`
or, more applicable in this case:
`color(white)("XXX")sin^2(A)=1-cos^2(A)`

Therefore the given
`color(white)("XXX")sin^2(A)=cos(A)-1`
can be written as
`color(white)("XXX")1-cos^2(A)=cos(A)-1`

Rearranging into standard quadratic form:
`color(white)("XXX")cos^2(A)+cos(A)-2=0`
then factoring:
`color(white)("XXX")(cos(A)+2)(cos(A)-1)=0`
giving:
#{:("either",cos(A)=-2," or",cos(A)=1), (,"impossible since "cos in[-1,+1],,rarrA=0+k*2pi"," k in ZZ) :}#