Match Sets of Trigonometric Equation Solutions in Intervals

To determine the correct option, we solve each trigonometric equation individually in the given intervals.

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Part (P)

We are given the equation: $\sin^6 x + \cos^4 x = 1, \quad x \in [-\pi, \pi]$

Using the identity $\cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x$, we rewrite the equation as: $\sin^6 x + 1 - 2\sin^2 x + \sin^4 x = 1$ $\sin^6 x + \sin^4 x - 2\sin^2 x = 0$

Let $y = \sin^2 x$. Since $y \ge 0$, we have: $y^3 + y^2 - 2y = 0 \implies y(y - 1)(y + 2) = 0$

Since $y = \sin^2 x \ge 0$, $y + 2 \neq 0$. Thus:

1. $y = 0 \implies \sin^2 x = 0 \implies x = k\pi$. For $x \in [-\pi, \pi]$, the solutions are: $x \in \{-\pi, 0, \pi\}$
2. $y = 1 \implies \sin^2 x = 1 \implies x = \pm \frac{\pi}{2}$.

Combining these, the set of solutions is $\left\{-\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi\right\}$. Thus, the number of elements is $5$.

$\implies \mathbf{(P) \rightarrow (5)}$

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Part (Q)

We are given the equation: $\sin^2 x + \cos^6 x = 1, \quad x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Using $\sin^2 x = 1 - \cos^2 x$, the equation becomes: $(1 - \cos^2 x) + \cos^6 x = 1$ $\cos^6 x - \cos^2 x = 0$ $\cos^2 x (\cos^4 x - 1) = 0$ $\cos^2 x (\cos^2 x - 1)(\cos^2 x + 1) = 0$

This gives two valid cases:

1. $\cos^2 x = 0 \implies x = \pm \frac{\pi}{2}$
2. $\cos^2 x = 1 \implies x = 0$

For $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the solutions are $x \in \left\{-\frac{\pi}{2}, 0, \frac{\pi}{2}\right\}$. Thus, the number of elements is $3$.

$\implies \mathbf{(Q) \rightarrow (3)}$

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Part (R)

We are given the equation: $\cos^2 \left(\frac{x}{2}\right) - \sin^2 x = \frac{1}{2}, \quad x \in [-\pi, \pi]$

Using the identities $\cos^2 \left(\frac{x}{2}\right) = \frac{1 + \cos x}{2}$ and $\sin^2 x = 1 - \cos^2 x$: $\frac{1 + \cos x}{2} - (1 - \cos^2 x) = \frac{1}{2}$ $\frac{1}{2} + \frac{1}{2}\cos x - 1 + \cos^2 x = \frac{1}{2}$ $\cos^2 x + \frac{1}{2}\cos x - 1 = 0$ $2\cos^2 x + \cos x - 2 = 0$

Solving for $\cos x$ using the quadratic formula: $\cos x = \frac{-1 \pm \sqrt{1 - 4(2)(-2)}}{2(2)} = \frac{-1 \pm \sqrt{17}}{4}$

Since $-1 \le \cos x \le 1$:

• $\cos x = \frac{\sqrt{17} - 1}{4} \approx 0.781 \in (-1, 1)$, which yields $2$ solutions in $x \in [-\pi, \pi]$.
• $\cos x = \frac{-\sqrt{17} - 1}{4} \approx -1.281 \notin [-1, 1]$, which yields no solutions.

Thus, the number of elements is $2$.

$\implies \mathbf{(R) \rightarrow (2)}$

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Part (S)

We are given the equation: $6 \sin^2 \left(\frac{x}{2}\right) - \cos 3x = 3, \quad x \in [-2\pi, 2\pi]$

Using the identity $2\sin^2\left(\frac{x}{2}\right) = 1 - \cos x$: $3(1 - \cos x) - \cos 3x = 3$ $3 - 3\cos x - \cos 3x = 3$ $\cos 3x + 3\cos x = 0$

Using the triple-angle identity $\cos 3x = 4\cos^3 x - 3\cos x$: $(4\cos^3 x - 3\cos x) + 3\cos x = 0$ $4\cos^3 x = 0 \implies \cos x = 0$

For $x \in [-2\pi, 2\pi]$, the solutions for $\cos x = 0$ are: $x \in \left\{-\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}\right\}$

Thus, the number of elements is $4$.

$\implies \mathbf{(S) \rightarrow (4)}$

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Conclusion

The matching is: $(P) \rightarrow (5),\ (Q) \rightarrow (3),\ (R) \rightarrow (2),\ (S) \rightarrow (4)$

This corresponds to Option B.