C-08-Q170mediumsingle_mcqIn 0<θ<π20 < \theta < \dfrac{\pi}{2}0<θ<2π, the equation 2cos2θ=1+2sin2θ2\cos^{2}\theta = 1 + 2\sin^{2}\theta2cos2θ=1+2sin2θ gives θ\thetaθ equal to —aπ6\dfrac{\pi}{6}6πbπ4\dfrac{\pi}{4}