The angular displacement α\alphaα (in degrees) of a robotic sensor arm during a precision sweep is governed by the relation:
−6cosα(tanαsinα−1)=7cosα−4 -6 \cos \alpha ( \tan \alpha \sin \alpha - 1 ) = 7 \cos \alpha - 4 −6cosα(tanαsinα−1)=7cosα−4Show that this relation can be simplified to the quadratic form:
6cos2α−cosα−2=0 6 \cos^2 \alpha - \cos \alpha - 2 = 0 6cos2α−cosα−2=0During a secondary calibration cycle, the arm operates such that the input angle is 3ϕ3\phi3ϕ. Find all values of ϕ \phi\,ϕ in the interval 0≤ϕ≤180∘ 0 \le \phi \le 180^\circ\,0≤ϕ≤180∘ such that:
−6cos3ϕ(tan3ϕsin3ϕ−1)=7cos3ϕ−4 -6 \cos 3\phi ( \tan 3\phi \sin 3\phi - 1 ) = 7 \cos 3\phi - 4 −6cos3ϕ(tan3ϕsin3ϕ−1)=7cos3ϕ−4giving your answers to one decimal place where appropriate.