cos(3pi/4 + x) + sin(2pi/4 - x) = 0

We will use trigonometric identities to solve.

We know that:

cos(x+y) = cosx*cosy - sinx*siny

==> cos(3pi/4+ x) = cos3pi/4*cosx - sin3pi/4*sinx

= -1/sqrt2*cosx - 1/sqrt2 * sinx

= -(cosx+sinx)/ sqrt2.............(1)

Also, we know that:

sin(x-y) = sinx*cosy - cosx*siny

==> sin(3pi/4 -x) = sin3pi/4*cosx - cos3pi/4*sinx

= 1/sqrt2 * cosx + 1/sqrt2 * sinx

= (cosx+sinx)/sqrt2................(2)

Now we will add (1) and (2).

==> cos(3pi/4+x)+sin(3pi/4-x) = -(cosx+sinx)/sqrt2 + (cosx+sinx)/sqrt2 = 0

==> cos(3pi/4+x)+sin(3pi/4-x) = 0 .............q.e.d

cos(3pi/4 +x)+sin (pi/4 -x) = 0.

We use cos(A+B) = cosA*cosB - sinA*sinB and sin (A-B) = sinAcosB- cosA*sinB.

Therefore

First term = cos(3pi/4 +x) = cos3pi/4*cosx - sin3pi/4 *sinx = (1sqrt2){-cosx-sinx}...(1), as cos pi/4 = -1/sqrt2 and sin 3pi/4 = 1/sqrt2.

2nd term = sin (3pi/4 -x) = sin3pi/4*cosx-cos3pi/4*sinx = (-1/sqrt2(cosx+sinx)...(2).

Therefore from (1) and (2) we get by adding:

cos(3pie/4 + x) + sin(3pie/4 - x) = 0.