Show that the equation 5−tanθcosθ=6cos2θ5 - \tan\theta \cos\theta = 6\cos^2\theta5−tanθcosθ=6cos2θ can be expressed in the form 6sin2θ−sinθ−1=06\sin^2 \theta - \sin \theta - 1 = 06sin2θ−sinθ−1=0
The diagram shows parts of the curves y=6cos2θy = 6\cos^2\thetay=6cos2θ and y=5−tanθcosθy = 5 - \tan\theta \cos\thetay=5−tanθcosθ, where θ \theta\,θ is in degrees. Solve the inequality 5−tanθcosθ>6cos2θ5 - \tan\theta \cos\theta > 6\cos^2\theta5−tanθcosθ>6cos2θ for 0∘≤θ<360∘0^\circ \leq \theta < 360^\circ0∘≤θ<360∘
