Express 2sinx−3cosx2\sin x - 3\cos x2sinx−3cosx in the form Rsin(x−α)R\sin(x - \alpha)Rsin(x−α), where R>0 R > 0\,R>0 and 0≤α≤π2\displaystyle 0 \leq \alpha \leq \frac{\pi}{2}0≤α≤2π. Give R R\,R in surd form and α \alpha\,α to three decimal places.
Hence find the greatest value of (2sinx−3cosx)2(2\sin x - 3\cos x)^2(2sinx−3cosx)2, and the smallest positive value of x x\,x at which this greatest value occurs.
Solve, for 0≤x≤2π0 \leq x \leq 2\pi0≤x≤2π, the equation
2sinx−3cosx=12\sin x - 3\cos x = 12sinx−3cosx=1
Give your answers to three decimal places.