By writing sin3θ \sin 3\theta\,sin3θ as sin(2θ+θ)\sin(2\theta + \theta)sin(2θ+θ) show that sin3θ=3sinθ−4sin3θ\sin 3\theta = 3\sin\theta - 4\sin^3\thetasin3θ=3sinθ−4sin3θ
Solve, for 0≤θ≤1800 \leq \theta \leq 1800≤θ≤180, the equation,
3sinθ−4sin3θ=0.4 3\sin\theta - 4\sin^3\theta = 0.4 3sinθ−4sin3θ=0.4Give your answers to 1 decimal place.