C-09-Q187mediumsingle_mcqSolve: 2sin2θ+3cosθ−3=02\sin^2\theta + 3\cos\theta - 3 = 02sin2θ+3cosθ−3=0, θ\thetaθ acute. θ=?\theta = ?θ=?a30∘30^\circ30∘b45∘45^\circ45∘c60ব্যাখ্যাSubstituting sin2θ=1−cos2θ\sin^2\theta=1-\cos^2\thetasin2θ=1−cos2θ gives 2cos2θ−3cosθ+1=0, i.e. . The non-zero acute solution is , so .