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≤1800 \leq x \leq 1800≤x≤180, the equation,
4cos3x−3cosx=−0.65 4\cos^3 x - 3\cos x = -0.65 4cos3x−3cosx=−0.65Give your answers to 1 decimal place.