Show that
6cos2θ+7sinθ−81−2sinθ≡3sinθ−2 \frac{6 \cos^2 \theta + 7 \sin \theta - 8}{1 - 2 \sin \theta} \equiv 3 \sin \theta - 2 1−2sinθ6cos2θ+7sinθ−8≡3sinθ−2Hence solve, for 0≤θ<360∘0 \leq \theta < 360^\circ0≤θ<360∘, the equation,
6cos2θ+7sinθ−81−2sinθ=2cosθ−2 \frac{6 \cos^2 \theta + 7 \sin \theta - 8}{1 - 2 \sin \theta} = 2 \cos \theta - 2 1−2sinθ6cos2θ+7sinθ−8=2cosθ−2
