Recall the angle sum identities:
[tex]\cos(a\pm b)=\cos a\cos b\mp\sin a\sin b[/tex]
[tex]\sin(a\pm b)=\sin a\cos b\pm\cos a\sin b[/tex]
So
[tex]\cos\left(\dfrac\pi4-x\right)=\dfrac1{\sqrt2}\cos x+\dfrac1{\sqrt2}\sin x[/tex]
[tex]\sin\left(\dfrac\pi4-x\right)=\dfrac1{\sqrt2}\cos x-\dfrac1{\sqrt2}\sin x[/tex]
Then the left hand side of the equation reduces significantly to give
[tex]\dfrac2{\sqrt2}\sin x=-1\implies\sin x=-\dfrac{\sqrt2}2=-\dfrac1{\sqrt2}[/tex]
In general, this occurs for [tex]x=-\dfrac\pi4+2n\pi[/tex] and [tex]x=-\dfrac{3\pi}4+2n\pi[/tex] where [tex]n[/tex] is any integer. We get solutions in the interval [tex][0,2\pi)[/tex] for [tex]n=1[/tex], for which we get
[tex]x=\dfrac{5\pi}4,\dfrac{7\pi}4[/tex]
