Question

\(1 + \cos x + \cos^2 x + \cos^3 x + \dots \text{ to } \infty = 4+2\sqrt{3}, \text{ then } x =\)

• A. \( n\pi \)
• B. \( (4n \pm 1)\frac{\pi}{3} \)
• C. \( (12n \pm 1)\frac{\pi}{6} \)
• D. \( (3n \pm 1)\frac{\pi}{3} \)

Hint:

Always rationalize expressions like \( \frac{1}{A+\sqrt{B}} \) to see if they simplify into recognizable trigonometric values.

Answer 1

The Correct Option is C

Step 1: Sum of Geometric Progression:

The series is an infinite GP with first term \( a=1 \) and common ratio \( r=\cos x \). For convergence, \( |\cos x|<1 \). Sum \( S = \frac{a}{1-r} = \frac{1}{1-\cos x} \).
Step 2: Solve for x:

\[ \frac{1}{1-\cos x} = 4 + 2\sqrt{3} \] \[ 1-\cos x = \frac{1}{4+2\sqrt{3}} \] Rationalize the denominator: \[ \frac{1(4-2\sqrt{3})}{(4+2\sqrt{3})(4-2\sqrt{3})} = \frac{4-2\sqrt{3}}{16 - 12} = \frac{4-2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2} \] So, \[ 1 - \cos x = 1 - \frac{\sqrt{3}}{2} \implies \cos x = \frac{\sqrt{3}}{2} \] The general solution for \( \cos x = \cos \frac{\pi}{6} \) is: \[ x = 2n\pi \pm \frac{\pi}{6} \] To match the options, find a common denominator: \[ x = \frac{12n\pi \pm \pi}{6} = (12n \pm 1)\frac{\pi}{6} \]
Step 4: Final Answer:

Option (C).