Let $$P = \{\theta : \sin\theta - \cos\theta = \sqrt{2}\cos\theta\}$$ and $$Q = \{\theta : \sin\theta + \cos\theta = \sqrt{2}\sin\theta\}$$, be two sets. Then

We begin with the first set

$$P=\{\theta:\;\sin\theta-\cos\theta=\sqrt2\cos\theta\}.$$

The defining equation can be rearranged step by step. First, we collect the terms containing $$\cos\theta$$ on the right‐hand side:

$$\sin\theta-\cos\theta=\sqrt2\cos\theta.$$

Adding $$\cos\theta$$ to both sides we obtain

$$\sin\theta=\cos\theta+\sqrt2\cos\theta.$$

Since the two terms on the right share the common factor $$\cos\theta$$, we factor it out:

$$\sin\theta=(1+\sqrt2)\cos\theta.$$

At this point we would like to divide by $$\cos\theta$$ in order to obtain a tangent. We therefore examine the possibility $$\cos\theta=0$$ separately. If $$\cos\theta=0$$, then $$\theta=\dfrac{\pi}{2}+k\pi,\;k\in\mathbb Z.$$ For such angles the left side of the original equation becomes

$$\sin\theta-\cos\theta=\pm1-0=\pm1,$$

while the right side is

$$\sqrt2\cos\theta=\sqrt2\cdot0=0.$$

The two sides are not equal, so no angle with $$\cos\theta=0$$ satisfies the equation. Hence $$\cos\theta\neq0$$ for every member of $$P$$, and division is safe. Dividing both sides by $$\cos\theta$$ gives

$$\dfrac{\sin\theta}{\cos\theta}=(1+\sqrt2).$$

By definition $$\dfrac{\sin\theta}{\cos\theta}=\tan\theta,$$ so we have

$$\tan\theta=\sqrt2+1.$$

The general solution of $$\tan\theta=m$$ is $$\theta=\arctan m+n\pi,\;n\in\mathbb Z$$. Therefore

$$P=\Bigl\{\theta:\;\theta=\arctan(\sqrt2+1)+n\pi,\;n\in\mathbb Z\Bigr\}.$$

Now we turn to the second set

$$Q=\{\theta:\;\sin\theta+\cos\theta=\sqrt2\sin\theta\}.$$

Again, we proceed step by step. We place the term in $$\sin\theta$$ on the right and the one in $$\cos\theta$$ on the left, obtaining

$$\cos\theta=\sqrt2\sin\theta-\sin\theta.$$

Factoring out $$\sin\theta$$ on the right yields

$$\cos\theta=(\sqrt2-1)\sin\theta.$$

Next we investigate whether $$\cos\theta=0$$ produces a solution. As before, $$\cos\theta=0$$ implies $$\theta=\dfrac{\pi}{2}+k\pi.$$ Substituting into the original equation gives

$$\sin\theta+\cos\theta=\pm1+0=\pm1,\qquad \sqrt2\sin\theta=\sqrt2(\pm1)=\pm\sqrt2,$$

and $$\pm1\neq\pm\sqrt2,$$ so no such angle is acceptable. Thus $$\cos\theta\neq0$$ for every angle in $$Q$$, and we may safely divide both sides of the equality $$\cos\theta=(\sqrt2-1)\sin\theta$$ by $$\cos\theta$$:

$$1=(\sqrt2-1)\dfrac{\sin\theta}{\cos\theta}.$$

Replacing $$\dfrac{\sin\theta}{\cos\theta}$$ with $$\tan\theta$$ gives

$$(\sqrt2-1)\tan\theta=1,$$ $$\tan\theta=\dfrac1{\sqrt2-1}.$$

To simplify $$\dfrac1{\sqrt2-1}$$ we rationalise the denominator:

$$\dfrac1{\sqrt2-1}\;=\;\dfrac{\sqrt2+1}{(\sqrt2-1)(\sqrt2+1)}\;=\;\dfrac{\sqrt2+1}{2-1}\;=\;\sqrt2+1.$$

Hence

$$\tan\theta=\sqrt2+1.$$

Using the same general tangent solution as before, we conclude

$$Q=\Bigl\{\theta:\;\theta=\arctan(\sqrt2+1)+n\pi,\;n\in\mathbb Z\Bigr\}.$$

The two descriptions of $$P$$ and $$Q$$ are identical, so

$$P=Q.$$

Among the given alternatives, Option C alone states this equality. No other option is consistent with the result.

Hence, the correct answer is Option 3.