Solve, for 0<θ<360∘0 < \theta < 360^\circ0<θ<360∘, the equation
4sin(θ+40∘)=3cos(θ+40∘) 4 \sin(\theta + 40^\circ) = 3 \cos(\theta + 40^\circ) 4sin(θ+40∘)=3cos(θ+40∘)giving your answers to one decimal place.
Show that the equation
2sin3x=6sinx−5sinxcosx 2 \sin^3 x = 6 \sin x - 5 \sin x \cos x 2sin3x=6sinx−5sinxcosxcan be written in the form
sinx(acos2x+bcosx+c)=0 \sin x (a \cos^2 x + b \cos x + c) = 0 sinx(acos2x+bcosx+c)=0where aaa, bbb and ccc are constants to be found.
Hence solve for −π≤x≤π-\pi \le x \le \pi−π≤x≤π the equation
2sin3x=6sinx−5sinxcosx 2 \sin^3 x = 6 \sin x - 5 \sin x \cos x 2sin3x=6sinx−5sinxcosxgiving your answers to two decimal places where appropriate.
