Question

If 2 sin θ + 3 cos θ = 2, then 3 sin θ – 2 cos θ = _________.

Answer 1

$$\begin{gathered} {\left(2\sin \theta +3\cos \theta \right)}^2+{\left(3\sin \theta -2\cos \theta \right)}^2 \\ =4{\sin }^2\theta +9{\cos }^2\theta +12\sin \theta \cos \theta +9{\sin }^2\theta +4{\cos }^2\theta -12\sin \theta \cos \theta \\ =13{\sin }^2\theta +13{\cos }^2\theta \\ =13\left({\sin }^2\theta +{\cos }^2\theta \right) \\ =13\left({\sin }^2\theta +{\cos }^2\theta =1\right) \end{gathered}$$
$$\begin{gathered} \therefore {\left(2\right)}^2+{\left(3\sin \theta -2\cos \theta \right)}^2=13\left(2\sin \theta +3\cos \theta =2\right) \\ \Rightarrow {\left(3\sin \theta -2\cos \theta \right)}^2=13-4=9 \\ \Rightarrow 3\sin \theta -2\cos \theta =\pm 3 \end{gathered}$$

If 2 sin θ + 3 cos θ = 2, then 3 sin θ – 2 cos θ = ____±3_____.