By writing cos3x \cos 3x\,cos3x as cos(2x+x)\cos(2x + x)cos(2x+x), show that cos3x=4cos3x−3cosx\cos 3x = 4\cos^3 x - 3\cos xcos3x=4cos3x−3cosx
Solve, for 0≤x≤π0 \leq x \leq \pi0≤x≤π, the equation,
4cos3x−3cosx=32 4\cos^3 x - 3\cos x = \frac{\sqrt{3}}{2} 4cos3x−3cosx=23Give your answers in terms of π\piπ.