Question

How do you solve `Sin(theta)^2 - cos(theta)^2 = sin(theta)`?

Answer 1

If we restrict ourselves to the interval `[0,2pi]`
then `theta in {pi/2, (7pi)/6,(11pi)/6}`

Explanation:

Remember (based on Pythagorean Theorem):
`color(white)("XXX")sin^2+cos^2=1color(white)("XX")rarrcolor(white)("XX")cos^2=1-sin^2`

There fore the given equation: `sin^2(theta)-cos^2(theta)=sin(theta)`
is equivalent to:
`color(white)("XXX")sin^2(theta)-(1-sin^2(theta))=sin(theta)`

`color(white)("XXX")2sin^2(theta)-sin(theta)-1=0`

Using the quadratic formula for roots:
`color(white)("XXX")sin(theta)=(1+-sqrt((-1)^2-4(2)(-1)))/(2(2)) = 1" or " -1/2`

Within the interval `theta in [0,2pi]`
using standard trigonometric ratios:

#{: ("if "sin(theta)=1,color(white)("XXXXXX"),"if "sin(theta)=-1/2), (color(white)("X")theta=pi/2,,color(white)("X")theta = (7pi)/6" or "(11pi)/6) :}#