The set of values of x for which `sinx cos^(3)x gt cos x sin^(3)x,0le x le pi` is

Options

A. 6
B. 7
C. 8
D. 9

Answer

To solve the inequality \( \sin x \cos^3 x > \cos x \sin^3 x \) for \( 0 \leq x \leq \pi \), we can start by rearranging the inequality.

### Step 1: Rearranging the Inequality
We can rewrite the inequality as:
\[ \sin x \cos^3 x - \cos x \sin^3 x > 0 \]

### Step 2: Factoring the Left Side
Notice that we can factor out \( \sin x \cos x \):
\[ \sin x \cos x (\cos^2 x - \sin^2 x) > 0 \]

### Step 3: Using the Identity
Recall that \( \cos^2 x - \sin^2 x = \cos 2x \). Thus, we can rewrite the inequality as:
\[ \sin x \cos x \cos 2x > 0 \]

### Step 4: Analyzing the Factors
Now we need to analyze when each factor is positive:
1. \( \sin x > 0 \)
2. \( \cos x > 0 \)
3. \( \cos 2x > 0 \)

### Step 5: Finding Intervals for Each Factor
- For \( \sin x > 0 \): This is true in the interval \( (0, \pi) \).
- For \( \cos x > 0 \): This is true in the interval \( (0, \frac{\pi}{2}) \).
- For \( \cos 2x > 0 \): This is true when \( 2x \) is in the intervals \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) or \( (2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}) \). For \( 0 \leq x \leq \pi \), we only consider \( 2x \in (0, \frac{\pi}{2}) \) which gives \( x \in (0, \frac{\pi}{4}) \) and \( 2x \in (\frac{3\pi}{2}, 2\pi) \) which gives \( x \in (\frac{3\pi}{4}, \pi) \).

### Step 6: Combining Intervals
Now we combine the intervals:
- From \( \sin x > 0 \): \( (0, \pi) \)
- From \( \cos x > 0 \): \( (0, \frac{\pi}{2}) \)
- From \( \cos 2x > 0 \): \( (0, \frac{\pi}{4}) \) and \( (\frac{3\pi}{4}, \pi) \)

Thus, the intervals where all conditions are satisfied are:
1. \( (0, \frac{\pi}{4}) \)
2. \( (\frac{3\pi}{4}, \pi) \)

### Final Step: Conclusion
The solution set for the inequality \( \sin x \cos^3 x > \cos x \sin^3 x \) in the interval \( 0 \leq x \leq \pi \) is:
\[ x \in \left(0, \frac{\pi}{4}\right) \cup \left(\frac{3\pi}{4}, \pi\right) \]