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

Question

3469 views around the world

You can reuse this answer
Creative Commons License

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.