Consider the following lists: List-I: (I$) \{ x$ in $[-2π/3, 2π/3] : \cos x + \sin x = 1 \}, ($II$) \{ x$ in $[-5π/18, 5π/18] : √3 \tan 3x = 1 \}, ($III$) \{ x$ in $[-6π/5, 6π/5] : 2 \cos(2x) = √3 \}, ($IV$) \{ x$ in $[-7π/4, 7π/4] : \sin x - \cos x = 1 \}$ List-II: (P) has two elements, (Q) has three elements, (R) has four elements, (S) has five elements, (T) has six elements The correct option is: (1) (I$)\to (P); ($II$)\to (S); ($III$)\to (P); ($IV$)\to (S)$ (2) (I$)\to (P); ($II$)\to (P); ($III$)\to (T); ($IV$)\to (R)$ (3) (I$)\to (Q); ($II$)\to (P); ($III$)\to (T); ($IV$)\to (S)$ (4) (I$)\to (Q); ($II$)\to (S); ($III$)\to (P); ($IV$)\to (R)$

Consider the following lists: | List-I | List-II | | --- | --- | | (I) { x ∈ [-2π/3, 2π/3] : cos x + sin x = 1 } | (P) has two elements | | (II) { x ∈ [-5π/18, 5π/18] : √3tan 3x = 1 } | (Q) has three elements | | (III) { x ∈ [-6π/5, 6π/5] : 2cos(2x) = √3 } | (R) has four elements | | (IV) { x ∈ [-7π/4, 7π/4] : si $\text{Given: }(\text{I})\ \cos x + \sin x = 1,\quad x \in \left[-\dfrac{2\pi}{3}, \dfrac{2\pi}{3}\right]$ $(\text{II})\ \sqrt{3}\tan 3x = 1,\quad x \in \left[-\dfrac{5\pi}{18}, \dfrac{5\pi}{18}\right]$ $(\text{III})\ 2\cos 2x = \sqrt{3},\quad x \in \left[-\dfrac{6\pi}{5}, \dfrac{6\pi}{5}\right]$ $(\text{IV})\ \sin x - \cos x = 1,\quad x \in \left[-\dfrac{7\pi}{4}, \dfrac{7\pi}{4}\right]$ $\text{To find: number of elements in each set } (\text{I})\text{–}(\text{IV})\text{, then match with } (\text{P})\text{–}(\text{T})$ $1.\ \text{reduce each equation to } \sin\theta = a,\ \cos\theta = a,\ \text{or } \tan\theta = a$ $2.\ \text{write the general solution for } \theta$ $3.\ \text{count solutions of } x \text{ inside the given interval}$ $\sin\theta = \sin\alpha \Rightarrow \theta = n\pi + (-1)^n\alpha$ $\cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha$ $\tan\theta = \tan\alpha \Rightarrow \theta = n\pi + \alpha$ $(\text{I})\ \cos x + \sin x = \sqrt{2}\sin\!\left(x + \dfrac{\pi}{4}\right) = 1$ $\sin\!\left(x + \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$ $x + \dfrac{\pi}{4} = \dfrac{\pi}{4} + 2k\pi \ \text{or}\ \dfrac{3\pi}{4} + 2k\pi$ $x = 2k\pi \quad \text{or} \quad x = \dfrac{\pi}{2} + 2k\pi$ $x = 0,\ \dfrac{\pi}{2} \in \left[-\dfrac{2\pi}{3}, \dfrac{2\pi}{3}\right]$ $(\text{I}) \to 2 \text{ elements} \to (\text{P})$ $(\text{II})\ \sqrt{3}\tan 3x = 1 \Rightarrow \tan 3x = \dfrac{1}{\sqrt{3}}$ $3x = \dfrac{\pi}{6} + k\pi \Rightarrow x = \dfrac{\pi}{18} + \dfrac{k\pi}{3}$ $x = \dfrac{\pi}{18},\ -\dfrac{5\pi}{18} \in \left[-\dfrac{5\pi}{18}, \dfrac{5\pi}{18}\right]$ $(\text{II}) \to 2 \text{ elements} \to (\text{P})$ $(\text{III})\ 2\cos 2x = \sqrt{3} \Rightarrow \cos 2x = \dfrac{\sqrt{3}}{2}$ $2x = 2k\pi \pm \dfrac{\pi}{6} \Rightarrow x = k\pi \pm \dfrac{\pi}{12}$ $x = \pm\dfrac{\pi}{12},\ \pm\dfrac{11\pi}{12},\ \pm\dfrac{13\pi}{12} \in \left[-\dfrac{6\pi}{5}, \dfrac{6\pi}{5}\right]$ $(\text{III}) \to 6 \text{ elements} \to (\text{T})$ $(\text{IV})\ \sin x - \cos x = \sqrt{2}\sin\!\left(x - \dfrac{\pi}{4}\right) = 1$ $\sin\!\left(x - \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$ $x - \dfrac{\pi}{4} = \dfrac{\pi}{4} + 2k\pi \ \text{or}\ \dfrac{3\pi}{4} + 2k\pi$ $x = \dfrac{\pi}{2} + 2k\pi \quad \text{or} \quad x = \pi + 2k\pi$ $x = -\dfrac{3\pi}{2},\ -\pi,\ \dfrac{\pi}{2},\ \pi \in \left[-\dfrac{7\pi}{4}, \dfrac{7\pi}{4}\right]$ $(\text{IV}) \to 4 \text{ elements} \to (\text{R})$ $(\text{I}) \to (\text{P}),\quad (\text{II}) \to (\text{P}),\quad (\text{III}) \to (\text{T}),\quad (\text{IV}) \to (\text{R})$ $(\text{1})\ \times,\quad (\text{2})\ \checkmark,\quad (\text{3})\ \times,\quad (\text{4})\ \times$ $\boxed{\text{Answer: Option (2)}}$