How do you solve cos^2 x = 1 + sinx in the interval (0, 2π)?

Use the identity that
sin²(x)+cos²(x)=1
so

cos²(x)=1+sin(x)
cos²(x)=sin²(x)+cos²(x)+sin(x)
minus cos²(x) both sides
0=sin²(x)+sin(x)
thefor
sin²(x)=-sin(x)
only possible if sin(x)=0
or if sin(x)=-1

that occurs at x=0 and x=3pi/2
but parethasees so 0 is not included
x=3pi/2

Sis less than or equal to 75 (bottom left)