Question

How to solve `cos^2(theta) = sin(theta) -1` in the domain of `0 ≤ theta ≤ 2pi` ?

Answer 1

Substitute `1 - sin^2(theta)` for `cos^2(theta)`
Solve the resulting quadratic.
Discard any extraneous roots.

Explanation:

Given: `cos^2(theta) = sin(theta) -1; 0 <= theta <= 2pi`

Substitute `1 - sin^2(theta)` for `cos^2(theta)`:

`1 - sin^2(theta) = sin(theta) -1; 0 <= theta <= 2pi`

Add `sin^2(theta)-1` to both sides of the equation:

`sin^2(theta) + sin(theta) -2=0; 0 <= theta <= 2pi`

This factors into:

`(sin(theta)+2)(sin(theta) -1)=0; 0 <= theta <= 2pi`

This can only be true when either factor is 0:

`(sin(theta)+2) = 0 and (sin(theta) -1)=0; 0 <= theta <= 2pi`

`sin(theta) = -2 and sin(theta)=-1; 0 <= theta <= 2pi`

We must discard the first solution, because it is outside of the the range of the sine function:

`sin(theta)=-1; 0 <= theta <= 2pi`

Take the inverse sine of of both sides:

`theta=sin^-1(-1); 0 <= theta <= 2pi`

This is a well know angle:

`theta=(3pi)/2`