Solve the equation `(sinx+cosx)^(1+sin2x)=2`, when `0 lexlepi`

Explanation

We know, `-sqrt(a^(2)+b^(2)) le a sintheta+bcostheta le sqrt(a^(2)+b^(2))` and `-1 le sintheta le1`
`therefore (sinx+cosx)` admits the maximum value as `sqrt(2)`.
and `(1+sin2x)` admits the maximum value as 2.
Also, `(sqrt(2))^(2)=2`.
`therefore` the equation could hold only when , `sinx + cosx= sqrt(2)` and `1+sin2x=2`
Now, `sinx+cosx=sqrt(2) rArr cos(x-pi/4)=1`
`rArr x=2npi+pi//4, n in I`...............(i)
and `1+sin2x=2 rArr sin2x=1-sinpi/2`
`rArr 2x=mpi +(-1)^(m)pi/2, m in I rArr 2x = mpi +(-1)^(m)pi/2, m in I rArr x=(mpi)/(2) + (-1)^(m)pi/4` ...........(ii)
The value of x in `[0,pi]` satisfying equations (i) and (ii) is `x=pi/4` (when n=0 and m=0)