Consider the following lists: List- I List- II (I) \ x ∈ [-...

Solving all questions one by one we get,

(i) $\left\{x \in \left[\dfrac{- 2\pi}{3}, \dfrac{2\pi}{3}\right], \cos x + \sin x = 1\right\}$

$\cos x + \sin x = 1$

$⇒ \sin \left(\dfrac{\pi}{4} + x\right) = \dfrac{1}{\sqrt{2}}$

$⇒ \dfrac{\pi}{4} + x = n\pi + {\left(- 1\right)}^{n}\dfrac{\pi}{4}$

$⇒ x = n\pi + {\left(- 1\right)}^{n}\dfrac{\pi}{4} - \dfrac{\pi}{4}$

So, $x \in \left\{0, \dfrac{\pi}{2}\right\}$

$∴ x$ has $2$ elements.$\to P$

(ii) \left\{x \in \left[\dfrac{- 5\pi}{18}, \dfrac{5\pi}{18}\right]: \sqrt{3}\tan 3x = 1\right\}$</p><p>Solving$\sqrt{3}\tan 3x = 1$</p><p>$⇒ \tan 3x = \dfrac{1}{\sqrt{3}}$</p><p>$⇒ 3x = n\pi + \dfrac{\pi}{6}$</p><p>$⇒ x = \dfrac{n\pi}{3} + \dfrac{\pi}{18}$</p><p>So,$x \in \left\{\dfrac{\pi}{18}, \dfrac{- 5\pi}{18}\right\}$</p><p>$∴ x$has$2$elements.$\to P$</p><p>(iii) \left\{x \in \left[\dfrac{- 6\pi}{5}, \dfrac{6\pi}{5}\right]: 2\cos 2x = \sqrt{3}\right\}$

$2\cos 2x = \sqrt{3}$

$⇒ \cos 2x = \dfrac{\sqrt{3}}{2}$

$⇒ 2x = 2n\pi \pm \dfrac{\pi}{6}$

$⇒ x = n\pi \pm \dfrac{\pi}{12}$

So, $x \in \left\{\pm \dfrac{\pi}{12}, \pi \pm \dfrac{\pi}{12}, - \pi \pm \dfrac{\pi}{12}\right\}$

$∴ x$ has $6$ elements.$\to T$

(iv) \left\{x \in \left[\dfrac{- 7\pi}{4}, \dfrac{7\pi}{4}\right]: \sin x - \cos x = 1\right\}$</p><p>$\sin x - \cos x = 1$</p><p>$⇒ \sin \left(x - \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$</p><p>$⇒ x - \dfrac{\pi}{4} = n\pi + {\left(- 1\right)}^{n}\dfrac{\pi}{4}$</p><p>$⇒ x = n\pi + {\left(- 1\right)}^{n}\dfrac{\pi}{4} + \dfrac{\pi}{4}$</p><p>So,$x \in \left\{\dfrac{\pi}{2}, \dfrac{- 3\pi}{2}, \pi, - \pi\right\}$</p><p>$∴ x$has$4$elements.$\to R$</p><p>$∴$option$B is correct.