$$\begin{aligned} & \text { Given equation, } \sin x-3 \sin 2 x+\sin 3 x=\cos x-3 \cos 2 x+\cos 3 x \\ & \Rightarrow \quad 2 \sin \left(\frac{x+3 x}{2}\right) \cdot \cos \left(\frac{3 x-x}{2}\right)-3 \sin 2 x \\ & =2 \cos \left(\frac{3 x+x}{2}\right) \cdot \cos \left(\frac{3 x-x}{2}\right)-3 \cos 2 x \end{aligned}$$

$$\begin{array}{rlrl} \Rightarrow & 2 \sin 2 x \cos x-3 \sin 2 x =2 \cos 2 x \cdot \cos x-3 \cos 2 x \\ \Rightarrow & \sin 2 x(2 \cos x-3) =\cos 2 x(2 \cos x-3) \\ \Rightarrow & \frac{\sin 2 x}{\cos 2 x} =1 \\ \Rightarrow & \tan 2 x =1 \\ \Rightarrow & \tan 2 x =\tan \frac{\pi}{4} \\ \Rightarrow & 2 x =n \pi+\frac{\pi}{4} \\ \therefore & x =\frac{n \pi}{2}+\frac{\pi}{8} \end{array}$$

Given equation is, $$(\sqrt{3}-1) \cos \theta+(\sqrt{3}+1) \sin \theta=2\quad\text{.... (i)}$$

Put $\quad\sqrt{3}-1=r \sin \alpha \quad \text { and } \quad \sqrt{3}+1=r \cos \alpha$

$$\begin{array}{llrl} \therefore & r^2 =(\sqrt{3}-1)^2+(\sqrt{3}+1)^2 \\ \Rightarrow & =3+1-2 \sqrt{3}+3+1+2 \sqrt{3} \\ \Rightarrow & r^2 =8 \\ \therefore & r =2 \sqrt{2} \\ & \text { now, } \tan \alpha =\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\tan \frac{\pi}{3}-\tan \frac{\pi}{4}}{1+\tan \frac{\pi}{3} \cdot \frac{\pi}{4}} \end{array}$$

$$\begin{array}{ll} \Rightarrow & \tan \alpha=\tan \left(\frac{\pi}{3}-\frac{\pi}{4}\right) \\ \Rightarrow & \tan \alpha=\tan \frac{\pi}{12} \\ \therefore & \alpha=\frac{\pi}{12} \end{array}$$

$$\begin{aligned} & \text { From Eq. (i), } r \sin \alpha \cos \theta+r \cos \alpha \sin \theta=2 \\ & \qquad \begin{aligned} r[\sin (\theta+\alpha)] & =2 \\ \Rightarrow \quad \sin (\theta+\alpha) & =\frac{2}{2 \sqrt{2}} \\ \Rightarrow \quad \sin (\theta+\alpha) & =\frac{1}{\sqrt{2}} \\ \Rightarrow \quad \sin (\theta+\alpha) & =\sin \frac{\pi}{4} \theta+\alpha=n \pi+(-1)^n \frac{\pi}{4} \\ \theta & =n \pi+(-1)^n \cdot \frac{\pi}{4}-\frac{\pi}{12} \end{aligned} \end{aligned}$$

Alternate Method

$$\begin{array}{lrl} (\sqrt{3}-1) \cos \theta+(\sqrt{3}+1) \sin \theta=2 \quad \text{... (i)}\\ \text { Put } \sqrt{3}-1 =r \cos \alpha \text { and } \sqrt{3}+1=r \sin \alpha \\ \therefore \quad r =2 \sqrt{2} \end{array}$$

Now,

$$\tan \alpha=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}$$

$$\begin{array}{ll} \Rightarrow & \tan \alpha=\frac{\tan \frac{\pi}{4}+\tan \frac{\pi}{6}}{1-\tan \frac{\pi}{4} \cdot \tan \frac{\pi}{6}} \\ \Rightarrow & \tan \alpha=\tan \left(\frac{\pi}{4}+\frac{\pi}{6}\right) \Rightarrow \tan \alpha=\tan \frac{5 \pi}{12} \\ \therefore & \quad \alpha=\frac{5 \pi}{12} \end{array}$$

$$\begin{aligned} & \text { From Eq. (i), } r \cos \alpha \cos \theta+r \sin \alpha \sin \theta=2 \\ & r[\cos (\theta-\alpha)]=2 \\ & \Rightarrow \quad \cos (\theta-\alpha)=\frac{2}{2 \sqrt{2}} \\ & \Rightarrow \quad \cos (\theta-\alpha)=\frac{1}{\sqrt{2}} \\ & \Rightarrow \quad \cos (\theta-\alpha)=\cos \frac{\pi}{4} \\ & \Rightarrow \quad \theta-\alpha=2 n \pi \pm \frac{\pi}{4} \\ \therefore \quad & \theta=2 n \pi \pm \frac{\pi}{4}+\frac{5 \pi}{12} \end{aligned}$$