1.Let θ∈(0,π2)\theta \in (0, \frac{\pi}{2})θ∈(0,2π). If the system of linear equations(1+cos2θ)x+sin2θ y+4sin3θ z=0,(1 + \cos^2 \theta) x + \sin^2 \theta \, y + 4 \sin 3\theta \, z = 0,(1+cos2θ)x+sin2θy+4sin3θz=0,cos2θ x+(1+sin2θ)y+4sin3θ z=0,\cos^2 \theta \, x + (1 + \sin^2 \theta) y + 4 \sin 3\theta \, z = 0,cos2θx+(1+sin2θ)y+4sin3θz=0,cos2θ x+sin2θ y+(1+4sin3θ)z=0\cos^2 \theta \, x + \sin^2 \theta \, y + (1 + 4 \sin 3\theta) z = 0cos2θx+sin2θy+(1+4sin3θ)z=0has a non-trivial solution, then the value of θ\thetaθ is:a.π4\dfrac{\pi}{4}4πb.5π8\dfrac{5\pi}{8}85πc.π2\dfrac{\pi}{2}2πd.π8\dfrac{\pi}{8}8πLogin to continueOnly logged in users canattempt or see the solution.
