Consider the equation. 1+sin2ax=cos‌x Now, 1+sin2ax≥1 And −1≤cos‌x≤1 It is possible only, 1+sin2ax=1 at x=2nπ Now, sin2ax=0 sin‌a‌x=0 ax=nπ a=
nπ
x
Solve further, a=
nπ
2nπ
=
1
2
But, a is irrational. It is possible only at x=0. Therefore, a is irrational only one solution satisfying the equation x=0