A) \[\left\{ \frac{\pi }{2},\frac{3\pi }{4} \right\}\]
B) \[\left\{ \frac{\pi }{4},\frac{3\pi }{4} \right\}\]
C) \[\left\{ \frac{\pi }{3},\frac{2\pi }{3} \right\}\]
D) \[\left\{ \frac{\pi }{6},\frac{5\pi }{6} \right\}\]
Correct Answer: C
Solution :
[c] We have \[{{8}^{1+|\cos x|+|\cos x{{|}^{2}}+|\cos x{{|}^{3}}+................to\,\,\infty \,\,}}={{4}^{3}}\] \[[\because \,\,{{\cos }^{2}}x=|{{\cos }^{2}}x|\] also \[|{{\cos }^{n}}x|=|\cos x{{|}^{n}}]\] \[\Rightarrow \,\,{{8}^{\frac{1}{1-|\cos x|}}}={{4}^{3}}\Rightarrow {{2}^{\frac{3}{1-|\cos x|}}}={{2}^{6}}\] \[\therefore \,\,\,\frac{3}{1-|\cos x|}=6\Rightarrow 1-|\cos x|=\frac{1}{2}\] \[\therefore \,\,|\cos x|=\frac{1}{2}\Rightarrow \cos x=\pm \frac{1}{2}\] \[\therefore \,\,x=\frac{\pi }{3},\frac{2\pi }{3}\]
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