Solving all question one by one we get,
(i) a m p ; { x ∈ [ − 2 π 3 , 2 π 3 ] , cos x + sin x = 1 } a m p ; cos x + sin x = 1 ⇒ a m p ; sin ( π 4 + x ) = 1 2 ⇒ a m p ; π 4 + x = n π + ( − 1 ) n π 4 \begin{aligned}\text { (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 \quad & \sin \left(\frac{\pi}{4}+x\right)=\frac{1}{\sqrt{2}} \\\\\Rightarrow & \frac{\pi}{4}+x=n \pi+(-1)^n \frac{\pi}{4}\end{aligned} (i) ⇒ ⇒ am p ; { x ∈ [ 3 − 2 π , 3 2 π ] , cos x + sin x = 1 } am p ; cos x + sin x = 1 am p ; sin ( 4 π + x ) = 2 1 am p ; 4 π + x = nπ + ( − 1 ) n 4 π a m p ; ⇒ x = n π + ( − 1 ) n π 4 − π 4 a m p ; So, x ∈ { 0 , π 2 } a m p ; ∴ x has 2 elements → P \begin{aligned}& \Rightarrow x=n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{4} \\\\& \text { So, } x \in\left\{0, \frac{\pi}{2}\right\} \\\\& \therefore x \text { has } 2 \text { elements } \rightarrow P\end{aligned} am p ; ⇒ x = nπ + ( − 1 ) n 4 π − 4 π am p ; So, x ∈ { 0 , 2 π } am p ; ∴ x has 2 elements → P a m p ; (ii) { x ∈ ( − 5 π 18 , 5 π 18 ) , 3 tan 3 x = 1 } a m p ; ⇒ tan 3 x = 1 3 a m p ; ⇒ x = n π + π 6 a m p ; ⇒ x = n π 3 + π 18 a m p ; So x ∈ { π 18 , − 5 π 18 } a m p ; ∴ x has 2 elements → P \begin{aligned}& \text { (ii) }\left\{x \in\left(\frac{-5 \pi}{18}, \frac{5 \pi}{18}\right), \sqrt{3} \tan 3 x=1\right\} \\\\& \Rightarrow \tan 3 x=\frac{1}{\sqrt{3}} \\\\& \Rightarrow x=n \pi+\frac{\pi}{6} \\\\& \Rightarrow x=\frac{n \pi}{3}+\frac{\pi}{18} \\\\& \text { So } x \in\left\{\frac{\pi}{18}, \frac{-5 \pi}{18}\right\} \\\\& \therefore x \text { has 2 elements } \rightarrow P\end{aligned} am p ; (ii) { x ∈ ( 18 − 5 π , 18 5 π ) , 3 tan 3 x = 1 } am p ; ⇒ tan 3 x = 3 1 am p ; ⇒ x = nπ + 6 π am p ; ⇒ x = 3 nπ + 18 π am p ; So x ∈ { 18 π , 18 − 5 π } am p ; ∴ x has 2 elements → P a m p ; (iii) { x ∈ [ − 6 π 5 , 6 π 5 ] , 2 cos 2 x = 3 } a m p ; ⇒ 2 cos 2 x = 3 a m p ; ⇒ cos 2 x = 3 2 a m p ; ⇒ 2 x = 2 n π ± π 6 a m p ; ⇒ x = n π ± π 12 a m p ; So x ∈ { ± π 12 , π ± π 12 , − π ± π 12 } a m p ; ∴ x has 6 elements → T \begin{aligned}& \text { (iii) }\left\{x \in\left[\frac{-6 \pi}{5}, \frac{6 \pi}{5}\right], 2 \cos 2 x=\sqrt{3}\right\} \\\\& \Rightarrow 2 \cos 2 x=\sqrt{3} \\\\& \Rightarrow \cos 2 x=\frac{\sqrt{3}}{2} \\\\& \Rightarrow 2 x=2 n \pi \pm \frac{\pi}{6} \\\\& \Rightarrow x=n \pi \pm \frac{\pi}{12} \\\\& \text { So } x \in\left\{ \pm \frac{\pi}{12}, \pi \pm \frac{\pi}{12},-\pi \pm \frac{\pi}{12}\right\} \\\\& \therefore x \text { has } 6 \text { elements } \rightarrow \mathrm{T}\end{aligned} am p ; (iii) { x ∈ [ 5 − 6 π , 5 6 π ] , 2 cos 2 x = 3 } am p ; ⇒ 2 cos 2 x = 3 am p ; ⇒ cos 2 x = 2 3 am p ; ⇒ 2 x = 2 nπ ± 6 π am p ; ⇒ x = nπ ± 12 π am p ; So x ∈ { ± 12 π , π ± 12 π , − π ± 12 π } am p ; ∴ x has 6 elements → T a m p ; (iv) { x ∈ [ − 7 π 4 , 7 π 4 ] , sin x − cos x = 1 } a m p ; ⇒ sin x − cos x = 1 a m p ; ⇒ sin ( x − π 4 ) = 1 2 a m p ; ⇒ x − π 4 = n π + ( − 1 ) n π 4 a m p ; ⇒ x = n π + ( − 1 ) n π 4 + π 4 a m p ; So, a m p ; x ∈ { π 2 , − 3 π 2 , π , − π } a m p ; ∴ x has 4 elements → R \begin{aligned}& \text { (iv) }\left\{x \in\left[\frac{-7 \pi}{4}, \frac{7 \pi}{4}\right], \sin x-\cos x=1\right\} \\\\& \Rightarrow \sin x-\cos x=1 \\\\& \Rightarrow \sin \left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \\\\& \Rightarrow x-\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{4} \\\\& \Rightarrow x=n \pi+(-1)^n \frac{\pi}{4}+\frac{\pi}{4} \\\\& \text { So, } \\\\& x \in\left\{\frac{\pi}{2}, \frac{-3 \pi}{2}, \pi,-\pi\right\} \\\\& \therefore x \text { has } 4 \text { elements } \rightarrow \mathrm{R}\end{aligned} am p ; (iv) { x ∈ [ 4 − 7 π , 4 7 π ] , sin x − cos x = 1 } am p ; ⇒ sin x − cos x = 1 am p ; ⇒ sin ( x − 4 π ) = 2 1 am p ; ⇒ x − 4 π = nπ + ( − 1 ) n 4 π am p ; ⇒ x = nπ + ( − 1 ) n 4 π + 4 π am p ; So, am p ; x ∈ { 2 π , 2 − 3 π , π , − π } am p ; ∴ x has 4 elements → R