A) \[-\frac{7\pi }{9}\]
B) \[-\frac{2\pi }{9}\]
C) \[0\]
D) \[\frac{5\pi }{9}\]
Correct Answer: C
Solution :
We have,\[\sqrt{3}\sec x+\text{cosec}\,x+2\,(\tan x-\cot x)=0\] |
\[\because \]\[\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}+2\left( \frac{\sin x}{\cos x}-\frac{\cos }{\sin x} \right)=0\]\[\Rightarrow \]\[\sqrt{3}\sin x+\cos x+2\,({{\sin }^{2}}x-{{\cos }^{2}}x)=0\]\[\Rightarrow \]\[\sqrt{3}\sin x+\cos x=2\cos 2x\] |
\[\Rightarrow \]\[\cos \left( x-\frac{\pi }{3} \right)=\cos 2x\]\[\Rightarrow \]\[\cos 2x=\cos \left( x-\frac{\pi }{3} \right)\]\[\Rightarrow \]\[2x=2n\pi \pm \left( x-\frac{\pi }{3} \right)\]\[\Rightarrow \]\[x=(6n-1)\frac{\pi }{3}\]or \[(6n+1)\frac{\pi }{9}\] |
\[\Rightarrow \]\[x=-\frac{\pi }{3},\frac{\pi }{9},\frac{7\pi }{9}\]and \[-\frac{5\pi }{9}\]in \[(-\pi \pi )\] |
\[\therefore \] Sum \[=\frac{-\pi }{3}+\frac{\pi }{9}+\frac{7\pi }{9}-\frac{5\pi }{9}=0\] |
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