Question

cos(x-pi/2)+sin^2x=0

Answer 1

\[\cos \left( x-\frac{ \pi }{2 } \right)=\cos x \cos \frac{ \pi }{ 2 }+\sin x \sin \frac{ \pi }{2 }=\cos x*0+\sin x*1=\sin x\]
\[\sin ^2x+\sin x=0,\sin x \left( \sin x+1 \right)=0,\sin x=0=\sin n \pi ,n \in z~or~I\]
\[x=n \pi \]
\[\sin x=-1=\sin \left( 2n \pi-\frac{ \pi }{ 2 } \right),x=2n \pi-\frac{ \pi }{2 }\]