Mathematics - Trigonometric Equation Question with Solution | TestHub

We have, $\frac{\sin \left(\lambda \alpha \right)}{\sin \alpha }-\frac{\cos \left(\lambda \alpha \right)}{\cos \alpha }=\lambda -1$

$\sin \left(\lambda \alpha \right)\cos \alpha -\cos \left(\lambda \alpha \right)\sin \alpha =\left(\lambda -1\right)\sin \alpha \cos \alpha$

$\sin \left(\lambda -1\right)\alpha =\left(\lambda -1\right)\sin \alpha \cos \alpha$

$\sin \left(\lambda -1\right)\alpha =\frac{\lambda -1}{2}\sin 2\alpha$

It is possible, when $\lambda -1=0\Rightarrow \lambda =1$

and $\frac{\lambda -1}{2}=1\Rightarrow \lambda =3$

$\therefore \lambda$ has only two solution.