Question
If sum of all the solutions of the equation $$8\cos x.\left( {\cos \left( {\frac{\pi }{6} + x} \right).\cos \left( {\frac{\pi }{6} - x} \right) - \frac{1}{2}} \right) - 1$$ in $$\left[ {0,\pi } \right]{\text{is }}k\pi ,$$ then $$k$$ is equal to:
A.
$$\frac{{13}}{9}$$
B.
$$\frac{{8}}{9}$$
C.
$$\frac{{20}}{9}$$
D.
$$\frac{{2}}{3}$$
Answer :
$$\frac{{13}}{9}$$
Solution :
$$\eqalign{
& \because \,\,8\cos x.\left( {{{\cos }^2}\frac{\pi }{6} - {{\sin }^2}x - \frac{1}{2}} \right) = 1 \cr
& \Rightarrow \,\,8\cos x\left( {\frac{3}{4} - \frac{1}{2} - {{\sin }^2}x} \right) = 1 \cr
& \Rightarrow \,\,8\cos x\left( {\frac{1}{4} - \left( {1 - {{\cos }^2}x} \right)} \right) = 1 \cr
& \Rightarrow \,\,8\cos x\left( {\frac{1}{4} - 1 + {{\cos }^2}x} \right) = 1 \cr
& \Rightarrow \,\,8\cos x\left( {{{\cos }^2}x - \frac{3}{4}} \right) = 1 \cr
& \Rightarrow \,\,8\left( {\frac{{4{{\cos }^3}x - 3\cos x}}{4}} \right) = 1 \cr
& \Rightarrow \,\,2\left( {4{{\cos }^3}x - 3\cos x} \right) = 1 \cr
& \Rightarrow \,\,2\cos 3x = 1 \cr
& \Rightarrow \,\,\cos 3x = \frac{1}{2} \cr
& \therefore \,\,3x = 2n\pi \pm \frac{\pi }{3},n \in 1 \cr
& \Rightarrow \,\,x = \frac{{2n\pi }}{3} \pm \frac{\pi }{9} \cr
& {\text{In }}x \in \left[ {0,\pi } \right]:x = \frac{\pi }{9},\frac{{2\pi }}{3} + \frac{\pi }{9},\frac{{2\pi }}{3} - \frac{\pi }{9},{\text{only}} \cr} $$
Sum of all the solutions of the equation
$$\eqalign{
& = \left( {\frac{1}{9} + \frac{2}{3} + \frac{1}{9} + \frac{2}{3} - \frac{1}{9}} \right)\pi \cr
& = \frac{{13}}{9}\pi \cr} $$