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

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