STD 11 - Trigonometrical equations — JEE STD 11 Science — Question

then $\sin \,(x + \lambda ) + \cos p\,(x + \lambda ) = \sin x + \cos px,\,\,\forall \,\,x \in R$

Putting $x = 0$ and replace $\lambda $ by $ - \lambda $, we have

$\sin \lambda + \cos p\lambda = 1$ and $ - \sin \lambda + \cos p\lambda = 1$

Solving these, we get $\sin \lambda = 0$ so $\lambda = n\pi $ and

$\cos p\lambda = 1$ so $p\lambda = 2m\pi .$

As $\lambda \ne 0,\,\,m$ and $n$ are non-zero integers.

Hence $p = \frac{{2m\pi }}{\lambda },$ which is rational.