MathematicsMedium244×since 2002Q4597Let θ∈(0,π2)\theta \in \left( {0,{\pi \over 2}} \right)θ∈(0,2π). If the system of linear equations (1+cos2θ)x+sin2θy+4sin3 θz=0(1 + {\cos ^2}\theta )x + {\sin ^2}\theta y + 4\sin 3\,\theta z = 0(1+cos2θ)x+sin2θy+4sin3θz=0 cos2θx+(1+sin2θ)y+4sin3 θz=0{\cos ^2}\theta x + (1 + {\sin ^2}\theta )y + 4\sin 3\,\theta z = 0cos2θx+(1+sin2θ)y+4sin3θz=0 cos2θx+sin2θy+(1+4sin3 θ)z=0{\cos ^2}\theta x + {\sin ^2}\theta y + (1 + 4\sin 3\,\theta )z = 0cos2θx+sin2θy+(1+4sin3θ)z=0 has a non-trivial solution, then the value of θ\thetaθ is :A4π9{{4\pi } \over 9}94πB7π18{{7\pi } \over {18}}187πCπ18{\pi \over {18}}18πD5π18{{5\pi } \over {18}}185πCheck answerSkip