Step 1: Understanding the Concept:
Trigonometric equations involving high powers or mixed multiple angles can often be solved by factorizing or using bounds (since \( \sin, \cos \in [-1, 1] \)). Step 3: Detailed Explanation:
1) Entry (P): \( \sin^6 x + \cos^4 x = 1 \).
\( \sin^6 x = 1 - \cos^4 x = (1-\cos^2 x)(1+\cos^2 x) = \sin^2 x (1+\cos^2 x) \).
Case 1: \( \sin x = 0 \implies x = 0, \pi, -\pi \) (3 solutions).
Case 2: \( \sin^4 x = 1 + \cos^2 x \).
Since \( \sin^4 x \le 1 \) and \( 1 + \cos^2 x \ge 1 \), this holds only when both are 1.
\( \sin^4 x = 1 \) and \( \cos^2 x = 0 \). This gives \( x = \pi/2, -\pi/2 \) (2 solutions).
Total: \( 3 + 2 = 5 \). (P) \(\to\) (5).
2) Entry (Q): \( \sin^2 x + \cos^6 x = 1 \implies \cos^6 x = 1 - \sin^2 x = \cos^2 x \).
\( \cos^2 x (\cos^4 x - 1) = 0 \).
Case 1: \( \cos x = 0 \implies x = \pi/2, -\pi/2 \).
Case 2: \( \cos x = \pm 1 \implies x = 0 \).
Total 3 solutions in \( [-\pi/2, \pi/2] \). (Q) \(\to\) (3).
3) Entry (R): \( \frac{1+\cos x}{2} - (1 - \cos^2 x) = 1/2 \).
\( 1 + \cos x - 2 + 2\cos^2 x = 1 \implies 2\cos^2 x + \cos x - 2 = 0 \).
\( \cos x = \frac{-1 \pm \sqrt{17}}{4} \). Only \( \frac{\sqrt{17}-1}{4} \in (0, 1) \) is valid.
This gives 2 solutions for \( x \) in \( [-\pi, \pi] \). (R) \(\to\) (2).
4) Entry (S): \( 3(1-\cos x) - \cos 3x = 3 \implies \cos 3x + 3\cos x = 0 \).
\( 4\cos^3 x - 3\cos x + 3\cos x = 0 \implies \cos x = 0 \).
In \( [-2\pi, 2\pi] \), solutions are \( \pm \pi/2, \pm 3\pi/2 \). Total 4. (S) \(\to\) (4). Step 4: Final Answer:
Matches (B).
