To solve this problem, we need to find the number of solutions for each equation in List-I within the specified interval and match it with the correct number of elements in List-II.

(I) Analysis

We need to solve the equation $\cos x+\sin x=1$ for $x \in\left[-\frac{2 \pi}{3}, \frac{2 \pi}{3}\right]$.

1. Rewrite the equation: We can rewrite $\cos x+\sin x$ in the form $R \cos(x - \alpha)$. Here $R = \sqrt{1^2 + 1^2} = \sqrt{2}$ and $\alpha = \frac{\pi}{4}$. The equation becomes: $\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right)=1$ $\sqrt{2}\left(\cos \frac{\pi}{4} \cos x+\sin \frac{\pi}{4} \sin x\right)=1$ $\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)=1$ $\cos \left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}$
2. Find the general solution: The general solution is $x-\frac{\pi}{4}=2 n \pi \pm \frac{\pi}{4}$, where $n$ is an integer. This gives two cases: Case 1: $x-\frac{\pi}{4}=2 n \pi+\frac{\pi}{4} \implies x=2 n \pi+\frac{\pi}{2}$ Case 2: $x-\frac{\pi}{4}=2 n \pi-\frac{\pi}{4} \implies x=2 n \pi$
3. Find solutions in the interval $\left[-\frac{2 \pi}{3}, \frac{2 \pi}{3}\right]$: The interval is approximately $[-2.09, 2.09] radians or $[-120^\circ, 120^\circ]$. From Case 1 ($x=2 n \pi+\frac{\pi}{2}$): For $n=0$, $x=\frac{\pi}{2}$. This is in the interval. From Case 2 ($x=2 n \pi$): For $n=0$, $x=0$. This is in the interval. Other integer values of $n$ give solutions outside the interval.
4. Conclusion: The set of solutions is $\{0, \frac{\pi}{2}\}$. It has two elements. Therefore, (I) maps to (P).

(II) Analysis

We need to solve $\sqrt{3} \tan 3 x=1$ for $x \in\left[-\frac{5 \pi}{18}, \frac{5 \pi}{18}\right]$.

1. Rewrite the equation: $\tan 3 x=\frac{1}{\sqrt{3}}=\tan \frac{\pi}{6}$
2. Find the general solution: The general solution is $3 x=n \pi+\frac{\pi}{6}$, where $n$ is an integer. $x=\frac{n \pi}{3}+\frac{\pi}{18}$
3. Find solutions in the interval $\left[-\frac{5 \pi}{18}, \frac{5 \pi}{18}\right]$: We have $-\frac{5 \pi}{18} \leq x \leq \frac{5 \pi}{18}$. Multiplying by 3 gives the interval for $3x$: $-\frac{5 \pi}{6} \leq 3x \leq \frac{5 \pi}{6}$. We need to find integer values of $n$ such that $-\frac{5 \pi}{6} \leq n \pi+\frac{\pi}{6} \leq \frac{5 \pi}{6}$. Subtracting $\frac{\pi}{6}$: $-\pi \leq n \pi \leq \frac{4 \pi}{6} = \frac{2 \pi}{3}$. Dividing by $\pi$: $-1 \leq n \leq \frac{2}{3}$. The possible integer values for $n$ are -1 and 0. For $n=-1$, $x=-\frac{\pi}{3}+\frac{\pi}{18}=\frac{-6\pi+\pi}{18}=-\frac{5\pi}{18}$. For $n=0$, $x=\frac{\pi}{18}$.
4. Conclusion: The set of solutions is $\{-\frac{5\pi}{18}, \frac{\pi}{18}\}$. It has two elements. Therefore, (II) maps to (P).

(III) Analysis

We need to solve $2 \cos (2 x)=\sqrt{3}$ for $x \in\left[-\frac{6 \pi}{5}, \frac{6 \pi}{5}\right]$.

1. Rewrite the equation: $\cos (2 x)=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}$
2. Find the general solution: The general solution for $2x$ is $2 x=2 n \pi \pm \frac{\pi}{6}$, where $n$ is an integer.
3. Find solutions in the interval $\left[-\frac{6 \pi}{5}, \frac{6 \pi}{5}\right]$: The interval for $x$ is $[-1.2\pi, 1.2\pi]$. The interval for $2x$ is $[-\frac{12 \pi}{5}, \frac{12 \pi}{5}] = [-2.4\pi, 2.4\pi]$. We find values of $n$ for which $2x$ lies in this interval. For $n=0$, $2x = \pm \frac{\pi}{6}$. These are valid. $x = \pm \frac{\pi}{12}$. For $n=1$, $2x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \approx 1.83\pi$ (valid) and $2x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6} \approx 2.17\pi$ (valid). So $x = \frac{11\pi}{12}, \frac{13\pi}{12}$. For $n=-1$, $2x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6}$ (valid) and $2x = -2\pi - \frac{\pi}{6} = -\frac{13\pi}{6}$ (valid). So $x = -\frac{11\pi}{12}, -\frac{13\pi}{12}$. For $n=2$, $2x = 4\pi - \frac{\pi}{6} = \frac{23\pi}{6} \approx 3.83\pi > 2.4\pi$ (invalid). For $n=-2$, $2x = -4\pi + \frac{\pi}{6} = -\frac{23\pi}{6} \approx -3.83\pi < -2.4\pi$ (invalid).
4. Conclusion: The solutions for $x$ are $$\left{\pm \frac{\pi}{12}, \pm \frac{11 \pi}{12}, \pm \frac{13 \pi}{12}\right}$. There are six elements. Therefore, (III) maps to (T).

(IV) Analysis

We need to solve $\sin x - \cos x=1$ for $x \in\left[-\frac{7 \pi}{4}, \frac{7 \pi}{4}\right]$.

1. Rewrite the equation: $\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x\right)=1$ $\sqrt{2}\left(\sin x \cos \frac{\pi}{4}-\cos x \sin \frac{\pi}{4}\right)=1$ $\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)=1$ $\sin \left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=\sin \frac{\pi}{4}$
2. Find the general solution: The general solution is $x-\frac{\pi}{4}=n \pi+(-1)^{n} \frac{\pi}{4}$. Case 1: $n$ is even ($n=2k$). $x-\frac{\pi}{4}=2 k \pi+\frac{\pi}{4} \implies x=2 k \pi+\frac{\pi}{2}$. Case 2: $n$ is odd ($n=2k+1$). $x-\frac{\pi}{4}=(2 k+1) \pi-\frac{\pi}{4} \implies x=(2 k+1) \pi$.
3. Find solutions in the interval $\left[-\frac{7 \pi}{4}, \frac{7 \pi}{4}\right] = [-1.75\pi, 1.75\pi]$: From Case 1 ($x=2 k \pi+\frac{\pi}{2}$): For $k=0$, $x=\frac{\pi}{2}$ (valid). For $k=-1$, $x=-2 \pi+\frac{\pi}{2}=-\frac{3 \pi}{2}=-1.5\pi$ (valid). From Case 2 ($x=(2 k+1) \pi$): For $k=0$, $x=\pi$ (valid). For $k=-1$, $x=-\pi$ (valid).
4. Conclusion: The set of solutions is $\{-\frac{3\pi}{2}, -\pi, \frac{\pi}{2}, \pi\}$. It has four elements. Therefore, (IV) maps to (R).

Summary of Matches

• (I) $\rightarrow$ (P) (2 elements)
• (II) $\rightarrow$ (P) (2 elements)
• (III) $\rightarrow$ (T) (6 elements)
• (IV) $\rightarrow$ (R) (4 elements)

Comparing this with the given options, the correct option is B.