Match each entry in List-I to the correct entry in List-II and...

For (P):
$\sin^6 x + \cos^4 x = 1$
$\sin^6 x + (1 - \sin^2 x)^2 = 1$
$\sin^6 x + 1 - 2\sin^2 x + \sin^4 x = 1$
$\sin^2 x (\sin^4 x + \sin^2 x - 2) = 0$
$\sin^2 x (\sin^2 x - 1)(\sin^2 x + 2) = 0$
This gives $\sin^2 x = 0 or$\sin^2 x = 1$.<br>In the interval$[-\pi, \pi]$, the solutions are$x \in \{-\pi, -\dfrac{\pi}{2}, 0, \dfrac{\pi}{2}, \pi\}$.<br>Number of solutions =$5$.<br><br>For (Q):<br>$\sin^2 x + \cos^6 x = 1$<br>$1 - \cos^2 x + \cos^6 x = 1$<br>$\cos^2 x (\cos^4 x - 1) = 0$<br>This gives$\cos^2 x = 0$or$\cos^2 x = 1$.<br>In the interval \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, the solutions are $x \in \left\{-\dfrac{\pi}{2}, 0, \dfrac{\pi}{2}\right\}$.
Number of solutions = $3$.

For (R):
$\cos^2\left(\dfrac{x}{2}\right) - \sin^2 x = \dfrac{1}{2}$
$\dfrac{1 + \cos x}{2} - (1 - \cos^2 x) = \dfrac{1}{2}$
$1 + \cos x - 2 + 2\cos^2 x = 1$
$2\cos^2 x + \cos x - 2 = 0$
Solving the quadratic equation for $\cos x$, we get:
$\cos x = \dfrac{-1 \pm \sqrt{17}}{4}$
Since $\cos x \in [-1, 1]$, we reject the negative root. Thus, $\cos x = \dfrac{\sqrt{17} - 1}{4}$.
Since $0 < \dfrac{\sqrt{17} - 1}{4} < 1$, there are exactly $2$ solutions in $[-\pi, \pi]$.
Number of solutions = $2$.

For (S):
$6\sin^2\left(\dfrac{x}{2}\right) - \cos 3x = 3$
$3(1 - \cos x) - (4\cos^3 x - 3\cos x) = 3$
$3 - 3\cos x - 4\cos^3 x + 3\cos x = 3$
$-4\cos^3 x = 0 \Rightarrow \cos x = 0$
In the interval $[-2\pi, 2\pi]$, the solutions are $x \in \left\{-\dfrac{3\pi}{2}, -\dfrac{\pi}{2}, \dfrac{\pi}{2}, \dfrac{3\pi}{2}\right\}$.
Number of solutions = $4$.

Therefore, the correct matching is (P) $\rightarrow$ (5), (Q) $\rightarrow$ (3), (R) $\rightarrow$ (2), (S) $\rightarrow$ (4).

Answer: (P) $\rightarrow$ (5), (Q) $\rightarrow$ (3), (R) $\rightarrow$ (2), (S) $\rightarrow$ (4)