Match the number of elements in List-I with the values in List-II. List-I: (P) The number of elements in the set $\{x$ in $[-\pi, \pi]: \sin^{6} x + \cos^{4} x = 1\}. (Q)$ The number of elements in the set $\{x$ in $[-\pi/2, \pi/2]: \sin^{2} x + \cos^{6} x = 1\}. (R)$ The number of elements in the set $\{x$ in $[-\pi, \pi]: \cos^{2}(x/2) - \sin^{2} x = \frac{1}{2}\}. (S)$ The number of elements in the set $\{x$ in $[-2pi, 2pi]: 6 \sin^{2}(x/2) - \cos 3x = 3\}.$ List-II: (1) is 1 (2) is 2 (3) is 3 (4) is 4 (5) is 5. Options$: (1) (P)\to (2), (Q)\to (5), (R)\to (3), (S)\to (4); (2) (P)\to (5), (Q)\to (3), (R)\to (2), (S)\to (4); (3) (P)\to (5), (Q)\to (4), (R)\to (1), (S)\to (3); (4) (P)\to (4), (Q)\to (3), (R)\to (2), (S)\to (5).$

| List-I | List-II | | --- | --- | | (P) The number of elements in the set {x ∈ [-π, π]: sin⁶x + cos⁴x = 1} | (1) is 1 | | (Q) The number of elements in the set {x ∈ [-π/2, π/2]: sin²x + cos⁶x = 1} | (2) is 2 | | (R) The number of elements in the set {x ∈ [-π, π]: cos²(x/2) - sin²x = 1/2} | (3) is 3 | | (S) The number of elements in the set {x ∈ [-2π, 2π]: 6sin²(x/2) - cos 3x = 3} | ( $(P)\ \sin^{6}x+\cos^{4}x=1,\quad x\in[-\pi,\pi]$ $(Q)\ \sin^{2}x+\cos^{6}x=1,\quad x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ $(R)\ \cos^{2}\left(\frac{x}{2}\right)-\sin^{2}x=\frac{1}{2},\quad x\in[-\pi,\pi]$ $(S)\ 6\sin^{2}\left(\frac{x}{2}\right)-\cos 3x=3,\quad x\in[-2\pi,2\pi]$ $\text{To find: the number of elements in each of }(P),(Q),(R),(S)$ $\text{then match each with a value from List-II }(1\text{ to }5)$ $\text{Step 1: write everything in ONE trig function}$ $\text{Step 2: solve the resulting equation}$ $\text{Step 3: count roots lying in the given interval}$ $\text{Let } u=\sin^{2}x \Rightarrow \cos^{2}x=1-u$ $\sin^{6}x=u^{3},\quad \cos^{4}x=(1-u)^{2}$ $u^{3}+(1-u)^{2}=1$ $u^{3}+1-2u+u^{2}=1$ $u^{3}+u^{2}-2u=0$ $u(u^{2}+u-2)=0$ $u^{2}+u-2=(u-1)(u+2)$ $u(u-1)(u+2)=0$ $u\in[0,1]\Rightarrow u=0\ \text{or}\ u=1$ $u=0:\ \sin x=0\Rightarrow x=-\pi,0,\pi\ (3)$ $u=1:\ \cos x=0\Rightarrow x=-\frac{\pi}{2},\frac{\pi}{2}\ (2)$ $(P):\ 3+2=5$ $\text{Let } v=\cos^{2}x \Rightarrow \sin^{2}x=1-v$ $\sin^{2}x+\cos^{6}x=1 \Rightarrow (1-v)+v^{3}=1$ $v^{3}-v=0$ $v(v^{2}-1)=0$ $v(v-1)(v+1)=0$ $v\in[0,1]\Rightarrow v=0\ \text{or}\ v=1$ $v=0:\ \cos x=0\Rightarrow x=-\frac{\pi}{2},\frac{\pi}{2}\ (2)$ $v=1:\ \sin x=0\Rightarrow x=0\ (1)$ $(Q):\ 2+1=3$ $\cos^{2}\frac{x}{2}=\frac{1+\cos x}{2}$ $\sin^{2}x=1-\cos^{2}x$ $\frac{1+\cos x}{2}-(1-\cos^{2}x)=\frac{1}{2}$ $1+\cos x-2+2\cos^{2}x=1$ $2\cos^{2}x+\cos x-1=1$ $2\cos^{2}x+\cos x-2=0$ $\text{Let } c=\cos x:\quad 2c^{2}+c-2=0$ $c=\frac{-1\pm\sqrt{17}}{4}$ $c=\frac{-1+\sqrt{17}}{4}\approx 0.78\in(-1,1)$ $c=\frac{-1-\sqrt{17}}{4}\approx -1.28<-1\ \text{(discard)}$ $(R):\ 2\ \text{roots}$ $\sin^{2}\frac{x}{2}=\frac{1-\cos x}{2}$ $6\sin^{2}\frac{x}{2}=3(1-\cos x)$ $3(1-\cos x)-\cos 3x=3$ $3-3\cos x-\cos 3x=3$ $-3\cos x-\cos 3x=0$ $\cos 3x=-3\cos x$ $\cos 3x=4\cos^{3}x-3\cos x$ $4\cos^{3}x-3\cos x=-3\cos x$ $4\cos^{3}x=0\Rightarrow\cos x=0$ $\cos x=0\Rightarrow x=\frac{\pi}{2}+n\pi$ $x\in[-2\pi,2\pi]:\ -\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$ $(S):\ 4$ $\begin{array}{|c|c|}\hline \text{Set} & \text{Elements}\\ \hline (P) & 5\\ (Q) & 3\\ (R) & 2\\ (S) & 4\\ \hline \end{array}$ $(P)\to 5,\ (Q)\to 3,\ (R)\to 2,\ (S)\to 4$ $(P)\to 5,\quad (Q)\to 3,\quad (R)\to 2,\quad (S)\to 4$ $\boxed{\text{Option }(2)}$