Match Sets of Angles Satisfying Trigonometric Equations within Given Intervals

To determine the correct matching option, we analyze each statement in List-I step-by-step.

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Entry (I)

We are given the set: $S_1 = \left\{x \in \left[-\frac{2\pi}{3}, \frac{2\pi}{3}\right] : \cos x + \sin x = 1\right\}$

Dividing the equation $\cos x + \sin x = 1$ by $\sqrt{2}$: $\frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x = \frac{1}{\sqrt{2}}$ $\cos\left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Thus, the general solution is: $x - \frac{\pi}{4} = 2k\pi \pm \frac{\pi}{4}, \quad k \in \mathbb{Z}$

This gives two cases:

1. $x = 2k\pi + \frac{\pi}{2}$
2. $x = 2k\pi$

Now we check for solutions within the interval $x \in \left[-\frac{2\pi}{3}, \frac{2\pi}{3}\right]$:

• For $k = 0$, $x = 0$ and $x = \frac{\pi}{2}$. Both lie inside $\left[-\frac{2\pi}{3}, \frac{2\pi}{3}\right]$.
• For any other integer $k$, $x$ falls outside this interval.

Thus, $S_1 = \left\{0, \frac{\pi}{2}\right\}$, which has two elements. $\text{(I)} \rightarrow \text{(P)}$

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Entry (II)

We are given the set: $S_2 = \left\{x \in \left[-\frac{5\pi}{18}, \frac{5\pi}{18}\right] : \sqrt{3} \tan 3x = 1\right\}$

Solving the equation: $\tan 3x = \frac{1}{\sqrt{3}}$ $3x = k\pi + \frac{\pi}{6} \implies x = \frac{k\pi}{3} + \frac{\pi}{18} = \frac{(6k+1)\pi}{18}, \quad k \in \mathbb{Z}$

We impose the interval constraint $x \in \left[-\frac{5\pi}{18}, \frac{5\pi}{18}\right]$: $-\frac{5\pi}{18} \le \frac{(6k+1)\pi}{18} \le \frac{5\pi}{18}$ $-5 \le 6k + 1 \le 5$ $-6 \le 6k \le 4 \implies -1 \le k \le \frac{2}{3}$

Since $k$ must be an integer, $k \in \{-1, 0\}$.

• For $k = -1$: $x = -\frac{5\pi}{18}$
• For $k = 0$: $x = \frac{\pi}{18}$

Thus, $S_2 = \left\{-\frac{5\pi}{18}, \frac{\pi}{18}\right\}$, which has two elements. $\text{(II)} \rightarrow \text{(P)}$

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Entry (III)

We are given the set: $S_3 = \left\{x \in \left[-\frac{6\pi}{5}, \frac{6\pi}{5}\right] : 2 \cos(2x) = \sqrt{3}\right\}$

Solving the equation: $\cos(2x) = \frac{\sqrt{3}}{2}$ $2x = 2k\pi \pm \frac{\pi}{6} \implies x = k\pi \pm \frac{\pi}{12}, \quad k \in \mathbb{Z}$

We find the values within the interval $\left[-\frac{6\pi}{5}, \frac{6\pi}{5}\right] = [-1.2\pi, 1.2\pi]$:

• For $k = 0$: $x = \pm \frac{\pi}{12} \approx \pm 0.0833\pi$ (2 solutions)
• For $k = 1$: $x = \pi + \frac{\pi}{12} = \frac{13\pi}{12} \approx 1.0833\pi$ and $x = \pi - \frac{\pi}{12} = \frac{11\pi}{12} \approx 0.9167\pi$ (2 solutions)
• For $k = -1$: $x = -\pi + \frac{\pi}{12} = -\frac{11\pi}{12} \approx -0.9167\pi$ and $x = -\pi - \frac{\pi}{12} = -\frac{13\pi}{12} \approx -1.0833\pi$ (2 solutions)

All $6$ values lie inside $[-1.2\pi, 1.2\pi]$.

Thus, $S_3 = \left\{-\frac{13\pi}{12}, -\frac{11\pi}{12}, -\frac{\pi}{12}, \frac{\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}\right\}$, which has six elements. $\text{(III)} \rightarrow \text{(T)}$

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Entry (IV)

We are given the set: $S_4 = \left\{x \in \left[-\frac{7\pi}{4}, \frac{7\pi}{4}\right] : \sin x - \cos x = 1\right\}$

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

The solutions for $x$ are: $x = 2k\pi + \frac{\pi}{2} \quad \text{or} \quad x = (2k+1)\pi, \quad k \in \mathbb{Z}$

Checking within the interval $\left[-\frac{7\pi}{4}, \frac{7\pi}{4}\right] = [-1.75\pi, 1.75\pi]$:

• From $x = 2k\pi + \frac{\pi}{2}$:
  • $k = 0 \implies x = \frac{\pi}{2}$
  • $k = -1 \implies x = -\frac{3\pi}{2}$
• From $x = (2k+1)\pi$:
  • $k = 0 \implies x = \pi$
  • $k = -1 \implies x = -\pi$

Thus, $S_4 = \left\{-\frac{3\pi}{2}, -\pi, \frac{\pi}{2}, \pi\right\}$, which has four elements. $\text{(IV)} \rightarrow \text{(R)}$

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Conclusion

Matching the entries:

• (I) $\rightarrow$ (P)
• (II) $\rightarrow$ (P)
• (III) $\rightarrow$ (T)
• (IV) $\rightarrow$ (R)

Hence, the correct option is B.