Show that the equation
4sin2x+9cosx−6=0 4 \sin^2 x + 9 \cos x - 6 = 0 4sin2x+9cosx−6=0can be written as
4cos2x−9cosx+2=0. 4 \cos^2 x - 9 \cos x + 2 = 0. 4cos2x−9cosx+2=0.Hence solve, for 0≤x<720∘0 \leq x < 720^\circ0≤x<720∘,
4sin2x+9cosx−6=0, 4 \sin^2 x + 9 \cos x - 6 = 0, 4sin2x+9cosx−6=0,giving your answers to 1 decimal place.
10 exam-style questions on Edexcel A Level Old Maths Trigonometry. Each one has a worked solution and a mark scheme showing where the marks go.