Solve 6sin2θ=cosθ+46 \sin^2 \theta = \cos \theta + 46sin2θ=cosθ+4 giving all the solutions for the interval 0≤θ<360∘0 \leq \theta < 360^\circ0≤θ<360∘
Hence, solve 6sin22θ=cos2θ+46 \sin^2 2\theta = \cos 2\theta + 46sin22θ=cos2θ+4 giving all the solutions for the interval 0≤θ<360∘0 \leq \theta < 360^\circ0≤θ<360∘