Answer:
[tex]\theta=0,\pi,\frac{3\pi}{2}[/tex] on the interval 0 ≤ θ <2π
Step-by-step explanation:
We have [tex]\sin \theta+1=\cos^2\theta[/tex]
We use the Pythagorean identity and substitute [tex]cos^2\theta=1-sin^2\theta[/tex]
[tex]\implies \sin \theta+1=1-\sin^2\theta[/tex]
[tex]\implies \sin \theta+1=1-\sin^2\theta[/tex]
[tex]\implies \sin^2\theta+\sin \theta=1-1[/tex]
[tex]\implies \sin^2\theta+\sin \theta=0[/tex]
Factor to get:
[tex]\implies \sin\theta(\sin \theta+1)=0[/tex]
[tex]\implies \sin\theta=0\:or\:\sin \theta=-1[/tex]
[tex]\theta=0,\pi,\frac{3\pi}{2}[/tex]