The number of real numbers λ for which the equality ( ( λα...

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

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

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

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

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

and $\dfrac{\lambda - 1}{2} = 1 ⇒ \lambda = 3$

$∴ \lambda$ has only two solution.