Trigonometric Functions & Equations

Concept:Solve trigonometric equations and count distinct solutions in given intervals.Explanation:For (P): Equation $\sin^6 x + \cos^4 x = 1$ in $[-\pi,\pi]$.Since $\sin^6 x \le \sin^2 x$ and $\cos^4 x \le \cos^2 x$, the sum is $\le 1$.Equality requires $\sin^6 x = \sin^2 x$ and $\cos^4 x = \cos^2 x$, so $\sin x \in \{-1,0,1\}$ and $\cos x \in \{-1,0,1\}$.Solutions: $x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi$ – total 5 elements. Hence (P) matches (5).For (Q): $\sin^2 x + \cos^6 x = 1$ in $[-\frac{\pi}{2},\frac{\pi}{2}]$.Rewrite as $1-\cos^2 x + \cos^6 x = 1 \Rightarrow \cos^2 x(\cos^4 x - 1) = 0$.Factor: $\cos^2 x(\cos^2 x - 1)(\cos^2 x + 1) = 0$; $\cos^2 x + 1 \neq 0$.So $\cos^2 x = 0$ gives $x = \pm\frac{\pi}{2}$; $\cos^2 x = 1$ gives $x = 0$.Thus 3 solutions: $-\frac{\pi}{2}, 0, \frac{\pi}{2}$. Hence (Q) matches (3).For (R): $\cos^2\frac{x}{2} - \sin^2 x = \frac12$ in $[-\pi,\pi]$.Use identities: $\cos^2\frac{x}{2} = \frac{1+\cos x}{2}$, $\sin^2 x = 1-\cos^2 x$.Equation becomes $\frac{1+\cos x}{2} - (1-\cos^2 x) = \frac12$.Simplify: $1+\cos x - 2 + 2\cos^2 x = 1 \Rightarrow 2\cos^2 x + \cos x - 2 = 0$.Solve: $\cos x = \frac{-1 \pm \sqrt{1+16}}{4} = \frac{-1 \pm \sqrt{17}}{4}$.Both values lie in $[-1,1]$, so each gives two solutions in $[-\pi,\pi]$? Actually check: $\frac{-1-\sqrt{17}}{4} \approx -1.28$ (less than -1) → invalid. Only $\frac{-1+\sqrt{17}}{4} \approx 0.78$ is valid, giving two solutions ($x = \pm \cos^{-1}(0.78)$) in $[-\pi,\pi]$. Hence (R) gives 2 elements. (Note: The existing solution says "Number of elements in $[-\pi,\pi]$ is 2".) So (R) matches (2).For (S): $6\sin^2\frac{x}{2} - \cos 3x = 3$ in $[-2\pi,2\pi]$.Use $\sin^2\frac{x}{2} = \frac{1-\cos x}{2}$: $6\cdot\frac{1-\cos x}{2} - \cos 3x = 3 \Rightarrow 3(1-\cos x) - \cos 3x = 3$.Simplify: $-3\cos x - \cos 3x = 0 \Rightarrow \cos 3x + 3\cos x = 0$.Use identity $\cos 3x = 4\cos^3 x - 3\cos x$: $(4\cos^3 x - 3\cos x) + 3\cos x = 0 \Rightarrow 4\cos^3 x = 0 \Rightarrow \cos x = 0$.In $[-2\pi,2\pi]$, $\cos x=0$ at $x = -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}$ – 4 solutions. Hence (S) matches (4).Thus correct matching: (P)→5, (Q)→3, (R)→2, (S)→4.Answer:Option B: (P)→(5), (Q)→(3), (R)→(2), (S)→(4).