Question

Solution set of inequality $\sin^3 x \cos x > \cos^3 x \sin x$, where $x \in (0, \pi)$, is

• A. $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
• B. $\left(\frac{3\pi}{4}, \pi\right)$
• C. $\left(0, \frac{\pi}{4}\right)$
• D. $\left(\frac{\pi}{2}, \frac{3\pi}{4}\right)$

Answer 1

Answer: (A), (B)

(A) $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ (B) $\left(\frac{3\pi}{4}, \pi\right)$

Answer 2

To solve the inequality $\sin^3 x \cos x > \cos^3 x \sin x$ for $x \in (0, \pi)$, we proceed as follows:

1. Rearrange the inequality: $\sin^3 x \cos x - \cos^3 x \sin x > 0$

2. Factor out common terms: $\sin x \cos x (\sin^2 x - \cos^2 x) > 0$

3. Apply trigonometric identities: We know that $\sin x \cos x = \frac{1}{2} \sin(2x)$ and $\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos(2x)$. Substitute these into the inequality: $\left(\frac{1}{2} \sin(2x)\right) (-\cos(2x)) > 0$ $-\frac{1}{2} \sin(2x) \cos(2x) > 0$

4. Simplify further using double angle identity: We know that $\sin(2A) \cos(2A) = \frac{1}{2} \sin(4A)$. Let $A = x$. So, $\sin(2x) \cos(2x) = \frac{1}{2} \sin(4x)$. Substitute this back: $-\frac{1}{2} \left(\frac{1}{2} \sin(4x)\right) > 0$ $-\frac{1}{4} \sin(4x) > 0$

5. Isolate $\sin(4x)$: Multiply by -4 (and reverse the inequality sign): $\sin(4x) < 0$

6. Determine the range for $4x$: The given domain for $x$ is $(0, \pi)$. Therefore, the domain for $4x$ is $(4 \times 0, 4 \times \pi)$, which is $(0, 4\pi)$.

7. Find intervals where $\sin(4x) < 0$ within $(0, 4\pi)$: The sine function is negative in the third and fourth quadrants. For $y \in (0, 4\pi)$, $\sin(y) < 0$ when:

   • $y \in (\pi, 2\pi)$
   • $y \in (3\pi, 4\pi)$

8. Substitute back $y = 4x$ and solve for $x$: Case 1: $\pi < 4x < 2\pi$ Divide by 4: $\frac{\pi}{4} < x < \frac{2\pi}{4}$ $\frac{\pi}{4} < x < \frac{\pi}{2}$

   Case 2: $3\pi < 4x < 4\pi$ Divide by 4: $\frac{3\pi}{4} < x < \frac{4\pi}{4}$ $\frac{3\pi}{4} < x < \pi$

9. Combine the solutions: The solution set for $x \in (0, \pi)$ is the union of these two intervals: $x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{4}, \pi\right)$.

Thus, the solution set is $\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{4}, \pi\right)$. Both options (A) and (B) represent parts of this solution.