For how many values of x does \frac{1}{p} become zero?
Consider the following for the next two (02) items that follow : Let p=\frac{1}{3}-\frac{\tan 3x}{\tan x} and q=1-3\tan^{2}x, 0 \lt x \lt \pi, x \neq \frac{\pi}{2}.
- A. No value
- B. Only one value
- C. Only two values ✓
- D. Only three values
Correct Answer: C. Only two values
Explanation
From the previous solution, p = \frac{-8}{3(1-3\tan^2 x)}. Therefore, the expression \frac{1}{p} = -\frac{3}{8}(1-3\tan^2 x). For this to be zero, \tan^2 x = \frac{1}{3} \implies \tan x = \pm \frac{1}{\sqrt{3}}. In the given interval 0 \lt x \lt \pi (excluding \pi/2), this gives x = \frac{\pi}{6} and x = \frac{5\pi}{6}. There are exactly two values.