Show that
sin3x≡3sinx−4sin3x \sin 3x \equiv 3 \sin x - 4 \sin^3 x sin3x≡3sinx−4sin3xHence, solve, for 0≤θ<π0 \leq \theta < \pi0≤θ<π,
8sin3x−6sinx+1=0 8 \sin^3 x - 6 \sin x + 1 = 0 8sin3x−6sinx+1=0137 exam-style questions on Edexcel A Level Maths Trigonometry and Modelling, covering 7.1 Addition Formulae, 7.2 Double Angle Formulae, 7.3 Solving Trigonometric Equations, 7.4 Simplifying a cos x +- b sin x, 7.5 Proving Trigonometric Identities, and 7.6 Modelling with Trigonometric Functions. Each one has a worked solution and a mark scheme showing where the marks go.