JEE Main 202118 Mar, Morning shiftMathematicsMediumMatrices and Determinants245× since 2002The solutions of the equation ∣1+sin2xsin2xsin2xcos2x1+cos2xcos2x4sin2x4sin2x1+4sin2x∣=0,(0<x<π)\left| {\matrix{ {1 + {{\sin }^2}x} & {{{\sin }^2}x} & {{{\sin }^2}x} \cr {{{\cos }^2}x} & {1 + {{\cos }^2}x} & {{{\cos }^2}x} \cr {4\sin 2x} & {4\sin 2x} & {1 + 4\sin 2x} \cr } } \right| = 0,(0 < x < \pi )1+sin2xcos2x4sin2xsin2x1+cos2x4sin2xsin2xcos2x1+4sin2x=0,(0<x<π), areAOption A. π12,π6{\pi \over {12}},{\pi \over 6}12π,6πBOption B. π6,5π6{\pi \over 6},{{5\pi } \over 6}6π,65πCOption C. 5π12,7π12{{5\pi } \over {12}},{{7\pi } \over {12}}125π,127πDOption D. 7π12,11π12{{7\pi } \over {12}},{{11\pi } \over {12}}127π,1211πCheckSkip · show solution
