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thanks for the help really appreciate it 

 Apr 27, 2021
 #1
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+3

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 :)

 Apr 27, 2021
edited by amygdaleon305  Apr 27, 2021
 #2
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+1

Thanks :)

Guest Apr 27, 2021

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