Understanding the Problem

The question asks for the number of values of \( x \) in the specific domain \( 0 < x < \pi \), with \( x \neq \frac{\pi}{2} \), for which the expression \( \frac{1}{p} \) becomes zero. The expression \( p \) is given by \( p=\frac{1}{3}-\frac{\tan 3 x}{\tan x} \).

Deriving the Expression for 1/p

For \( \frac{1}{p} \) to be zero, the quantity \( p \) must be such that its reciprocal is zero. This typically happens when the numerator of \( \frac{1}{p} \) is zero and the denominator is non-zero. Let's first express \( \frac{1}{p} \) as a single fraction.

The expression for \( p \) is:

\( p = \frac{1}{3} - \frac{\tan 3x}{\tan x} \)

We can rewrite \( \tan x \) and \( \tan 3x \) in terms of sine and cosine:

\( p = \frac{1}{3} - \frac{\sin 3x / \cos 3x}{\sin x / \cos x} \)

\( p = \frac{1}{3} - \frac{\sin 3x \cos x}{\cos 3x \sin x} \)

To combine these terms, we find a common denominator, which is \( 3 \cos 3x \sin x \):

\( p = \frac{(\cos 3x \sin x) \times 1}{3 \cos 3x \sin x} - \frac{(\sin 3x \cos x) \times 3}{3 \cos 3x \sin x} \)

\( p = \frac{\cos 3x \sin x - 3 \sin 3x \cos x}{3 \cos 3x \sin x} \)

Now, to find \( \frac{1}{p} \), we take the reciprocal of this expression:

\( \frac{1}{p} = \frac{3 \cos 3x \sin x}{\cos 3x \sin x - 3 \sin 3x \cos x} \)

Condition for \( \frac{1}{p} \) to be Zero

For a fraction \( \frac{A}{B} \) to be equal to zero, the numerator \( A \) must be zero and the denominator \( B \) must be non-zero. In our case, for \( \frac{1}{p} = 0 \), we need the numerator \( 3 \cos 3x \sin x \) to be zero and the denominator \( \cos 3x \sin x - 3 \sin 3x \cos x \) to be non-zero.

Setting the Numerator to Zero

The numerator is \( 3 \cos 3x \sin x \). Setting it equal to zero gives:

\( 3 \cos 3x \sin x = 0 \)

This equation is satisfied if either \( \cos 3x = 0 \) or \( \sin x = 0 \).

Case 1: \( \sin x = 0 \)

We are looking for solutions in the domain \( 0 < x < \pi \). In this open interval, the equation \( \sin x = 0 \) has no solutions. The values where \( \sin x = 0 \) are \( x = 0, \pi, 2\pi, \dots \), which are not included in the strictly open interval \( (0, \pi) \).

Case 2: \( \cos 3x = 0 \)

The general solution for \( \cos \theta = 0 \) is \( \theta = (n + \frac{1}{2})\pi \), where \( n \) is an integer. So, for \( \cos 3x = 0 \), we have:

\( 3x = (n + \frac{1}{2})\pi = \frac{(2n+1)\pi}{2} \)

Solving for \( x \), we get:

\( x = \frac{(2n+1)\pi}{6} \)

We need to find the integer values of \( n \) for which \( x \) falls within the domain \( 0 < x < \pi \), excluding \( x = \frac{\pi}{2} \).

• For \( n = 0 \): \( x = \frac{(2(0)+1)\pi}{6} = \frac{\pi}{6} \). This value is between 0 and \( \pi \), and \( \frac{\pi}{6} \neq \frac{\pi}{2} \). So, \( x = \frac{\pi}{6} \) is a potential solution.
• For \( n = 1 \): \( x = \frac{(2(1)+1)\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \). This value is between 0 and \( \pi \), but it is excluded by the domain condition \( x \neq \frac{\pi}{2} \). So, \( x = \frac{\pi}{2} \) is not a solution.
• For \( n = 2 \): \( x = \frac{(2(2)+1)\pi}{6} = \frac{5\pi}{6} \). This value is between 0 and \( \pi \), and \( \frac{5\pi}{6} \neq \frac{\pi}{2} \). So, \( x = \frac{5\pi}{6} \) is a potential solution.
• For \( n = 3 \): \( x = \frac{(2(3)+1)\pi}{6} = \frac{7\pi}{6} \). This value is greater than \( \pi \), so it is outside the domain \( 0 < x < \pi \).
• For any integer \( n < 0 \), the value of \( x \) will be less than 0, which is outside the domain.

Thus, the potential values of \( x \) from the numerator condition within the domain are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).

Checking the Denominator Condition

The denominator of \( \frac{1}{p} \) is \( D = \cos 3x \sin x - 3 \sin 3x \cos x \). For \( \frac{1}{p} \) to be zero, this denominator must be non-zero at the potential solution values \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).

Checking at \( x = \frac{\pi}{6} \)

If \( x = \frac{\pi}{6} \), then \( 3x = 3 \cdot \frac{\pi}{6} = \frac{\pi}{2} \). We need to evaluate \( D = \cos(\frac{\pi}{2}) \sin(\frac{\pi}{6}) - 3 \sin(\frac{\pi}{2}) \cos(\frac{\pi}{6}) \). Using known values: \( \cos(\frac{\pi}{2}) = 0 \), \( \sin(\frac{\pi}{6}) = \frac{1}{2} \), \( \sin(\frac{\pi}{2}) = 1 \), \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \). \( D = (0)(\frac{1}{2}) - 3(1)(\frac{\sqrt{3}}{2}) = 0 - \frac{3\sqrt{3}}{2} = -\frac{3\sqrt{3}}{2} \). Since \( -\frac{3\sqrt{3}}{2} \neq 0 \), the denominator is non-zero at \( x = \frac{\pi}{6} \).

Checking at \( x = \frac{5\pi}{6} \)

If \( x = \frac{5\pi}{6} \), then \( 3x = 3 \cdot \frac{5\pi}{6} = \frac{5\pi}{2} \). Note that \( \frac{5\pi}{2} = 2\pi + \frac{\pi}{2} \).

We need to evaluate \( D = \cos(\frac{5\pi}{2}) \sin(\frac{5\pi}{6}) - 3 \sin(\frac{5\pi}{2}) \cos(\frac{5\pi}{6}) \). Using known values: \( \cos(\frac{5\pi}{2}) = \cos(\frac{\pi}{2}) = 0 \), \( \sin(\frac{5\pi}{2}) = \sin(\frac{\pi}{2}) = 1 \). \( \sin(\frac{5\pi}{6}) = \sin(\pi - \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2} \). \( \cos(\frac{5\pi}{6}) = \cos(\pi - \frac{\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2} \). Substitute these values into the denominator expression: \( D = (0)(\frac{1}{2}) - 3(1)(-\frac{\sqrt{3}}{2}) = 0 - (-\frac{3\sqrt{3}}{2}) = \frac{3\sqrt{3}}{2} \). Since \( \frac{3\sqrt{3}}{2} \neq 0 \), the denominator is non-zero at \( x = \frac{5\pi}{6} \).

Conclusion

Both \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \) satisfy the condition that the numerator of \( \frac{1}{p} \) is zero and the denominator is non-zero. Both these values are within the specified domain \( 0 < x < \pi \), \( x \neq \frac{\pi}{2} \).

Thus, there are exactly two values of \( x \) in the given domain for which \( \frac{1}{p} \) becomes zero.