+526 We have,
\(sin(x+{\pi \over 2})-cos(x-\pi)=1\)
⇒\(cosx-cos(x-\pi)=1\)
Now, we know the formula \(cosx-cosy=-2sin{x+y\over 2}sin{x-y \over 2}\)
⇒\(-2sin{x+x-\pi\over 2}sin{x-x+\pi\over 2}=1\)
⇒\(-2sin({x-{\pi\over 2}})sin{\pi\over 2}=1\)
⇒\(2sin({\pi\over 2}-x)sin{\pi\over 2}=1\)
⇒\(2cosxsin{\pi\over 2}=1\)
⇒\(cosx=1/2\)
⇒\(x={\pi\over 3}\)
~Hope this helps :)
