Question

If the sum of all the solutions of the equation $8\cos \left(x\right)\begin{array}{cl}\left(\cos \left(\left(\frac{\mathrm{\pi }}{6}\right)+x\right)\right).& \left(\cos \left(\left(\frac{\pi }{6}\right)-\left(x\right)\right)-\frac{1}{2}\right)\end{array}=1\mathrm{in}\left[0,\mathrm{\pi }\right]\mathrm{is}\mathrm{k\pi }$, then $k$ is equal to

• A. $\frac{8}{9}$
• B. $\frac{20}{9}$
• C. $\frac{2}{3}$
• D. $\frac{13}{9}$

Answer 1

The correct option is D

$\frac{13}{9}$

Explanation for the correct option:

Step 1: Solve the given equation

$\begin{array}{rcl}& \Rightarrow & 8\cos x\left(\cos \left(\left(\frac{\mathrm{\pi }}{6}\right)+x\right)\right).\left(\cos \left(\left(\frac{\pi }{6}\right)-\left(x\right)\right)-\frac{1}{2}\right)=1\\ & \Rightarrow & 8.\cos x\left[\frac{\left(\cos \left(\frac{\pi }{3}\right)+\cos 2x-1\right)}{2}\right]=1\\ & \Rightarrow & 4\cos x\left(\cos 2x-\left(\frac{1}{2}\right)\right)=1\\ & \Rightarrow & 4\cos x\left(2{\cos }^{2}x-\left(\frac{3}{2}\right)\right)=1\\ & \Rightarrow & 8{\cos }^{3}x-6\cos x-1=0\\ & \Rightarrow & 2(4{\cos }^{3^{}}x-3\cos x)-1=0\\ & \Rightarrow & 2\cos 3x-1=0\end{array}$

Step 2: Find all of the possible x values.

$\begin{array}{rcl}& & \begin{array}{l}\therefore \cos \left(3x\right)=\frac{1}{2}\\ \Rightarrow \cos \left(3x\right)=\cos \left(\frac{\pi }{3}\right)\\ \Rightarrow 3x=2n\pi \pm \frac{\pi }{3}\\ \Rightarrow x=\frac{2n\pi }{3}\pm \frac{\pi }{9}\\ \Rightarrow x=\frac{7\pi }{9},\frac{5\pi }{9},\frac{\pi }{9},\text{in}[0,\pi ]\end{array}\end{array}$

Therefore, $k=\frac{13}{9}$

Hence, the correct answer is an option (D).