Consider the following lists :

List-I	List-II
(I) $\left\{x\in \left[-\frac{2\pi }{3},\frac{2\pi }{3}\right]:\cos x+\sin x=1\right\}$	(P) has two elements
(II) $\left\{x\in \left[-\frac{5\pi }{18},\frac{5\pi }{18}\right]:\sqrt{3}\tan 3x=1\right\}$	(Q) has three elements
(III) $\left\{x\in \left[-\frac{6\pi }{5},\frac{6\pi }{5}\right]:2\cos \left(2x\right)=\sqrt{3}\right\}$	(R) has four elements
(IV) $\left\{x\in \left[-\frac{7\pi }{4},\frac{7\pi }{4}\right]:\sin x-\cos x=1\right\}$	(S) has five elements
	(T) has six elements

The correct option is :

The correct option is (I) → (P); (II) → (P); (III) → (T); (IV) → (R).

Let us analyze each item in List-I one by one and count the number of elements (solutions) in the given intervals.

Item (I):
We are given the set:

$\left\{x\in \left[-\frac{2\pi }{3},\frac{2\pi }{3}\right]:\cos x+\sin x=1\right\}$

Rewrite the trigonometric equation $\cos x+\sin x=1$ by dividing both sides by $\sqrt{2}$:

$\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x=\frac{1}{\sqrt{2}}$

$\cos \left(x-\frac{\pi }{4}\right)=\cos \left(\frac{\pi }{4}\right)$

Thus, $x-\frac{\pi }{4}=2k\pi \pm \frac{\pi }{4}$ for $k\in \mathbb{Z}$.
This gives two cases for $x$:
1) $x=2k\pi +\frac{\pi }{2}$
2) $x=2k\pi$

Now, we find solutions lying in the interval $\left[-\frac{2\pi }{3},\frac{2\pi }{3}\right]$:
- For $x=2k\pi$: taking $k=0$ gives $x=0$ (since $0\in \left[-\frac{2\pi }{3},\frac{2\pi }{3}\right]$).
- For $x=2k\pi +\frac{\pi }{2}$: taking $k=0$ gives $x=\frac{\pi }{2}$ (since $\frac{\pi }{2}=\frac{3\pi }{6}\le \frac{4\pi }{6}=\frac{2\pi }{3}$).

Therefore, the set has two elements ($x=0,\frac{\pi }{2}$).
Hence, (I) → (P).

Item (II):
We are given the set:

$\left\{x\in \left[-\frac{5\pi }{18},\frac{5\pi }{18}\right]:\sqrt{3}\tan 3x=1\right\}$

Solving the equation $\tan 3x=\frac{1}{\sqrt{3}}$:

$3x=k\pi +\frac{\pi }{6}$

$x=\frac{k\pi }{3}+\frac{\pi }{18}=\frac{(6k+1)\pi }{18}$

Given $x\in \left[-\frac{5\pi }{18},\frac{5\pi }{18}\right]$, we test integer values for $k$:
- For $k=0$: $x=\frac{\pi }{18}$, which is in the range.
- For $k=-1$: $x=\frac{-5\pi }{18}$, which is in the range.
- For $k=1$: $x=\frac{7\pi }{18}$, which lies outside the range.

Therefore, the set has two elements ($x=-\frac{5\pi }{18},\frac{\pi }{18}$).
Hence, (II) → (P).

Item (III):
We are given the set:

$\left\{x\in \left[-\frac{6\pi }{5},\frac{6\pi }{5}\right]:2\cos \left(2x\right)=\sqrt{3}\right\}$

Solving $\cos \left(2x\right)=\frac{\sqrt{3}}{2}$:

$2x=2k\pi \pm \frac{\pi }{6}$

$x=k\pi \pm \frac{\pi }{12}$

We convert the domain limits to have a common denominator of 12:
$\left[-\frac{6\pi }{5},\frac{6\pi }{5}\right]=\left[-\frac{14.4\pi }{12},\frac{14.4\pi }{12}\right]$

Now list the valid values of $x$:
- For $k=0$: $x=\frac{\pi }{12},-\frac{\pi }{12}$ (2 solutions)
- For $k=1$: $x=\pi +\frac{\pi }{12}=\frac{13\pi }{12}$ and $x=\pi -\frac{\pi }{12}=\frac{11\pi }{12}$ (2 solutions)
- For $k=-1$: $x=-\pi +\frac{\pi }{12}=-\frac{11\pi }{12}$ and $x=-\pi -\frac{\pi }{12}=-\frac{13\pi }{12}$ (2 solutions)

All 6 of these solutions fall within $[-14.4\backslash pi/12,\; 14.4\backslash pi/12]$.
Therefore, the set has six elements.
Hence, (III) → (T).

Item (IV):
We are given the set:

$\left\{x\in \left[-\frac{7\pi }{4},\frac{7\pi }{4}\right]:\sin x-\cos x=1\right\}$

Dividing both sides of $\sin x-\cos x=1$ by $\sqrt{2}$:

$\sin \left(x-\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}=\sin \left(\frac{\pi }{4}\right)$

General solutions for $x$ are:
1) $x-\frac{\pi }{4}=2k\pi +\frac{\pi }{4}\Rightarrow x=2k\pi +\frac{\pi }{2}$
2) $x-\frac{\pi }{4}=2k\pi +\pi -\frac{\pi }{4}\Rightarrow x=2k\pi +\pi$

Check for values of $x$ in the interval $\left[-\frac{7\pi }{4},\frac{7\pi }{4}\right]$:
- From $x=2k\pi +\frac{\pi }{2}$:
- $k=0\Rightarrow x=\frac{\pi }{2}$
- $k=-1\Rightarrow x=-\frac{3\pi }{2}$ (which is $-\frac{6\pi }{4}\in \left[-\frac{7\pi }{4},\frac{7\pi }{4}\right]$)
- From $x=2k\pi +\pi$:
- $k=0\Rightarrow x=\pi$
- $k=-1\Rightarrow x=-\pi$

Thus, there are four elements ($x=\frac{\pi }{2},-\frac{3\pi }{2},\pi ,-\pi$).
Hence, (IV) → (R).

Combining all matches:
(I) → (P), (II) → (P), (III) → (T), (IV) → (R)