TS EAMCET 5 May 2018 Shift 2 Solved Paper

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