Question

If (2 sin θ + 3 cos θ) = 2, show that (3 sin θ − 2 cos θ) = ± 3.

Answer 1

$$\begin{gathered} \text{Given,}(2\sin \theta +3\cos \theta )=2...\left(\text{i}\right) \\ \begin{array}{l}\text{We have}{(2\sin \theta +3\cos \theta )}^2+{(3\sin \theta -2\cos \theta )}^2\\ \end{array} \\ {\text{=4sin}}^2\theta +9{\cos }^2\theta +12\sin \theta \cos \theta +9{\sin }^2\theta +4{\cos }^2\theta -12\sin \theta \cos \theta \\ \text{=4}({\sin }^2\theta +{\cos }^2\theta )+9({\sin }^2\theta +{\cos }^2\theta ) \\ \text{=4+9} \\ \text{=13} \\ \mathrm{i}.\mathrm{e}.,{(2\sin \theta +3\cos \theta )}^2+{(3\sin \theta -2\cos \theta )}^2=13 \\ =>2^2+{(3\sin \theta -2\cos \theta )}^2=13 \\ =>{(3\sin \theta -2\cos \theta )}^2=13-4 \\ =>{(3\sin \theta -2\cos \theta )}^2=9 \\ =>(3\sin \theta -2\cos \theta )=\pm 3 \end{gathered}$$