A) \[15\pi \]
B) \[30\pi \]
C) \[\frac{110\pi }{3}\]
D) None of these
Correct Answer: B
Solution :
We have,\[\cos x\cdot \cos \left( \frac{\pi }{3}+x \right)\cos \left( \frac{\pi }{3}-x \right)=\frac{1}{4}\] \[\Rightarrow \] \[\cos x\left( \frac{1}{4}{{\cos }^{2}}x-\frac{3}{4}{{\sin }^{2}}x \right)=\frac{1}{4}\] or \[\frac{\cos x}{4}(4{{\cos }^{2}}x-3)=\frac{1}{4}\] \[\Rightarrow \] \[4{{\cos }^{3}}x-3\cos x=1\] \[(\because \,\,\cos 3x=4{{\cos }^{3}}x-3\cos x)\] or \[\cos 3x=1\] \[\Rightarrow \] \[3x=2n\pi \] \[\Rightarrow \]\[x=\frac{2n\pi }{3},\] where\[n=0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9\] \[(\because \,\,x\in [0,\,\,6\pi ])\] \[\therefore \]The required sum\[=\frac{2\pi }{3}\sum\limits_{n=0}^{9}{n=30\pi }\]
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