A) \[\frac{8}{9}\]
B) \[\frac{20}{9}\]
C) \[\frac{2}{3}\]
D) \[\frac{13}{9}\]
Correct Answer: D
Solution :
\[8\cos x\cdot \left( \cos \left( \frac{\pi }{6}+x \right)\cdot \cos \left( \frac{\pi }{6}-x \right)-\frac{1}{2} \right)=1\] \[8\cos x\left[ {{\cos }^{2}}x-{{\sin }^{2}}\frac{\pi }{6}-\frac{1}{2} \right]\] \[8\cos x\left[ {{\cos }^{2}}x-\frac{3}{4} \right]=1\] \[8{{\cos }^{3}}x-6\cos x=1\] \[{{\cos }^{3}}x=\frac{1}{2}\] \[3x=\frac{\pi }{3},2\pi -\frac{\pi }{3},2\pi +\frac{\pi }{3}\] \[x=\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9}\] Sum\[=\frac{13\pi }{9}\]
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