Show that sin3x≡3sinx−4sin3x\sin 3x \equiv 3\sin x - 4\sin^3 xsin3x≡3sinx−4sin3x.
Hence solve, for 0≤x<π0 \leq x < \pi0≤x<π, the equation
8sin3x−6sinx+1=08\sin^3 x - 6\sin x + 1 = 08sin3x−6sinx+1=0
Give your answers in terms of π\piπ.
221 exam-style questions on CCEA A Level Maths 3.4 Trigonometry (A-level only), covering 3.4.1 Trigonometry (A-level only), 3.4.2 Trigonometry (A-level only), 3.4.3 Trigonometry (A-level only), 3.4.4 Trigonometry (A-level only), 3.4.5 Trigonometry (A-level only), 3.4.6 Trigonometry (A-level only), 3.4.7 Trigonometry (A-level only), and 3.4.8 Trigonometry (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.