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∘c60∘60^\circ60∘d90∘90^\circ90∘ব্যাখ্যাSubstituting sin2θ=1−cos2θ\sin^2\theta=1-\cos^2\thetasin2θ=1−cos2θ gives 2cos2θ−3cosθ+1=02\cos^2\theta-3\cos\theta+1=02cos2θ−3cosθ+1=0, i.e. (2cosθ−1)(cosθ−1)=0(2\cos\theta-1)(\cos\theta-1)=0(2cosθ−1)(cosθ−1)=0. The non-zero acute solution is cosθ=12\cos\theta=\tfrac12cosθ=21, so θ=60∘\theta=60^\circθ=60∘.