Question

find x where 0 ≤ x ≤ \\(\frac{\pi}{2}\\).
3(tan x + cos x) = \\(\sqrt{3}\\) + 3 cos x
\\(\frac{\pi}{?}\\)

Explanation:

Step1: Simplify the equation

Start with the given equation: \( 3(\tan x + \cos x) = \sqrt{3} + 3\cos x \).
Distribute the 3 on the left side: \( 3\tan x + 3\cos x = \sqrt{3} + 3\cos x \).
Subtract \( 3\cos x \) from both sides: \( 3\tan x = \sqrt{3} \).

Step2: Solve for \(\tan x\)

Divide both sides by 3: \( \tan x = \frac{\sqrt{3}}{3} \).

Step3: Find \(x\) in the interval \([0, \frac{\pi}{2}]\)

We know that \( \tan \frac{\pi}{6} = \frac{\sqrt{3}}{3} \), and since \( x \) is in the interval \([0, \frac{\pi}{2}]\), the solution is \( x = \frac{\pi}{6} \). So the denominator in the fraction \(\frac{\pi}{?}\) is 6.

Answer:

6