Question

If sum of all the solutions of the equation \( 8 \cos x \). \( \left(\cos \left(\frac{\pi}{6}+x\right) \cdot \cos \left(\frac{\pi}{6}-x\right)-\frac{1}{2}\right)-1 \) in \( [0, \pi] \) is \( k \pi \), then \( k \) is equal to : (a) \( \frac{13}{9} \) (b) \( \frac{8}{9} \) (c) \( \frac{20}{9} \) (d) \( \frac{2}{3} \)

Comments

No equation is given. It is algebraic expression.
Wrong data.

Answer 1

8 cos x [cos(π/6 ​+ x)cos(π/6 ​− x) −1/2​] = 1

using 2 cosA cosB = cos (A+B) + cos (A−B), we get

8 cosx [cos (π/3​) + cos(2x)​/2 − 1/2​] = 1

∴ 4 cosx [1/2 ​+ cos(2x) − 1] = 1

∴ 4 cosx [cos(2x)−1/2​] = 1

∴ 4 cosx cos(2x) − 2 cosx = 1

using 2 cosA cosB = cos(A+B) + cos(A−B), we get

2 (cos3x + cosx) − 2 cosx = 1

∴ cos3x = 1/2​

∴ 3x = 2nπ ± π/3​

∴ x = 2nπ/3 ​± π/9​

Solutions in [0,π] are π/9​, 2π/3 ​− π/9​, 2π/3 ​+ π/9​

Hence, their sum = π/9 ​+ 5π/9 ​+7π/9 ​= 13π/9​

∴ k = 13/9​

Option A is Right Answer