I've solved it by taking modulus of each side. Is there any other possible way to solve this question? If $\left(a,b,c,d\right) \in \mathbb{R}^4$ and if $$a+bi=\frac{c}{d+\cos(x)+i\sin(x)}$$

Show that $$(d^2-1)(a^2+b^2) +c^2 = 2acd$$
cross multiplication leads to the equation$$ad+a\cos(x)-b\sin(x)+i(bd+b\cos(x)+a\sin(x))=c$$
your equation is true
writing
$$a\cos(x)-b\sin(x)=-ad$$
and 
$$a\sin(x)+b\cos(x)=-bd$$
squaring both equations and adding we get
$$a^2(\sin^2(x)+\cos^2(x))+b^2(\sin^2(x)+\cos^2(x))=a^2d^2+b^2d^2$$
can you finish?
from this equation we get
$$a^2+b^2=d^2(a^2+b^2)$$ thus $$d^2=1$$ if $$a^2+b^2\neq 0$$
cross multiplication leads to the equation$$ad+a\cos(x)-b\sin(x)+i(bd+b\cos(x)+a\sin(x))=c$$
your equation is true
writing
$$a\cos(x)-b\sin(x)=-ad$$
and 
$$a\sin(x)+b\cos(x)=-bd$$
squaring both equations and adding we get
$$a^2(\sin^2(x)+\cos^2(x))+b^2(\sin^2(x)+\cos^2(x))=a^2d^2+b^2d^2$$
can you finish?
from this equation we get
$$a^2+b^2=d^2(a^2+b^2)$$ thus $$d^2=1$$ if $$a^2+b^2\neq 0$$