How do you solve $Sin(x+pi)+cos(x-(pi/2))=1$ ?

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Answer 1 · 2016-02-14T16:41:14

$x$ doesn't exist.

$[1]" "sin(x+pi)+cos(x-pi/2)=1$

$[2]" "sin(x+pi)+cos(-(pi/2-x))=1$

Identity: $cos(-a)=cosa$

$[3]" "sin(a+pi)+cos(pi/2-a)=1$

Identity: $cos(pi/2-a)=sina$

$[4]" "sin(x+pi)+sinx=1$

Identity: $sin(a+b)=sinacosb+cosasinb$

$[5]" "(sinxcospi+cosxsinpi)+sinx=1$

$[6]" "(-1)sinx+(0)cosx+sinx=1$

$[7]" "-sinx+sinx=1$

$[8]" "0!=1$

This means that $x$ doesn't exist.

How do you solve Sin(x+pi)+cos(x-(pi/2))=1Sin(x+pi)+cos(x-(pi/2))=1?