Write 9sin(x) + 12 cos(x) in the form Rsin(x+y) and hence solve 9sin(x) + 12 cos(x) = 3

9sin(x) + 12 cos(x) = Rsin(x+y) =R(sin(x)cos(y)+cos(x)sin(y))= (Rcos(y))sin(x) + (Rsin(y))cos(x)Therefore by matching the coefficientsRsin(y)=12, Rcos(y)=9 [1]SoRsin(y)/Rcos(y) = 12/9 = 4/3, therefore tan(y) = 4/3So y =tan^-1(4/3)=0.927Now we solve for RR^2 = 12^2 + 9^2 = 225 [by drawing the triangle or by squaring and adding the equations in [1]]So R=159sin(x) + 12 cos(x) = 15 sin(x+0.927)Now we take the equation given and can rewrite it as15 sin(x+0.927) = 3So sin(x+0.927) = 3/15 =1/5X+0.927 = sin^-1(0.2) = 0.20136, 2.94023, 9.22342 [we discount the first result as wouldn’t give x in the desired range]So x= 2.01, 8.30