The answers are
a. 0
b. π/6
c. 5π/6
d. π
Explanation:
If we replace each of these options into the equation we can see that all makes the equation true:
[tex]\begin{gathered} \sin ^20-\sin 0+1=\cos ^20 \\ 0-0+1=1 \\ 1=1\text{ true} \end{gathered}[/tex][tex]\begin{gathered} \sin ^2\frac{\pi}{6}-\sin \frac{\pi}{6}+1=\cos ^2\frac{\pi}{6} \\ \frac{1}{2^2}-\frac{1}{2}+1=(\frac{\sqrt[]{3}}{2})^2 \\ \frac{1}{4}-\frac{1}{2}+1=\frac{3}{4} \\ 1-\frac{1}{4}=\frac{3}{4} \\ \frac{3}{4}=\frac{3}{4}\text{ true} \end{gathered}[/tex][tex]\begin{gathered} \sin ^2\frac{5\pi}{6}-\sin \frac{5\pi}{6}+1=\cos ^2\frac{5\pi}{6} \\ \frac{1}{2^2}-\frac{1}{2}+1=(-\frac{\sqrt[]{3}}{2})^2^{} \\ \frac{1}{4}-\frac{1}{2}+1=\frac{3}{4} \\ \frac{3}{4}=\frac{3}{4}\text{ true} \end{gathered}[/tex][tex]\begin{gathered} \sin ^2\pi-\sin \pi+1=\cos ^2\pi \\ 0-0+1=(-1)^2 \\ 1=1\text{ true} \end{gathered}[/tex]