Answer:
First option [tex]0, \frac\pi2, \pi, \frac{3\pi}2[/tex]
Step-by-step explanation:
first of all, let's replace the lhs with the equivalent in sine (from the pythagoran identiy on sine and cosine)
[tex]1-sin^2\theta =1-sin\theta[/tex]
At this point we take 1 from both side, and change signs just out of convenience.
[tex]\sin^2\theta =sin\theta[/tex]
Now we have 2 options. If [tex]sin \theta =0 \leftrightarrow \theta = k\pi[/tex] the equation is verified, and we got two solutions [tex]\theta =0; \theta = \pi[/tex]. Else, we can divide by [tex]sin\theta[/tex] and we get
[tex]sin\theta = 1 \leftrightarrow \theta = \frac\pi2+k\pi[/tex].
We can add everything together and write
[tex]\theta = k\frac\pi2; k\in\{0,1,2,3\}[/tex] or, in our case, the first answer.
