Question #aa66a

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Answer 1 · 2017-11-17T06:24:11

Write this as follows

#sinx-cosx=1#

Multiply with #sqrt2/2# both sides

#sqrt2/2*(sinx-cosx)=sqrt2/2#

#sin(x-pi/4)=sin(pi/4)#

Hence the general solutions are

#x-pi/4=2*κ*pi+pi/4# and #x-pi/4=2*k*pi+3pi/4#

or

#χ=2*k*pi+pi/2# and #x=2*k*pi+pi#

Because x belongs in # (0,2*pi]# acceptable solutions are

#x=pi# and #x=pi/2#