Consider the following lists: List- I List- II I x ∈ - 2 π 3 , 2 π 3 : cos x + sin x = 1 P…

Solving all questions one by one we get,

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

$\cos x+\sin x=1$

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

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

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

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

$\therefore x$ has $2$ elements.$\to P$

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

Solving $\sqrt{3}\tan 3x=1$

$\Rightarrow \tan 3x=\frac{1}{\sqrt{3}}$

$\Rightarrow 3x=n\pi +\frac{\pi }{6}$

$\Rightarrow x=\frac{n\pi }{3}+\frac{\pi }{18}$

So, $x\in \left\{\frac{\mathrm{\pi }}{18},\frac{-5\mathrm{\pi }}{18}\right\}$

$\therefore x$ has $2$ elements.    $\to P$

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

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

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

$\Rightarrow 2x=2n\pi \pm \frac{\pi }{6}$

$\Rightarrow x=n\pi \pm \frac{\pi }{12}$

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

$\therefore x$ has $6$ elements.$\to T$

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

$\sin x-\cos x=1$

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

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

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

So, $x\in \left\{\frac{\mathrm{\pi }}{2},\frac{-3\mathrm{\pi }}{2},\mathrm{\pi },-\mathrm{\pi }\right\}$

$\therefore x$ has $4$ elements.     $\to R$

$\therefore$ option $B$ is correct.