Solve the equation: `cos^2[pi/4(sinx+sqrt(2)cos^2x)]-tan^2[x+pi/4tan^2x]=1`

Explanation

`cos^(2) [pi/4(sin x+sqrt(2) cos^(2) x)]-tan^(2) [x+pi/4 tan^(2) x]=1`
`rArr sin^(2) [pi/4 (sin x+ sqrt(2) cos^(2) x)]+tan^(2) [x+pi/4 tan^(2) x]=0`
It is possible only when
`sin^(2) [pi/4 (sin x+sqrt(2) cos^(2) x)]=0`...(i)
and `tan^(2) [x+pi/4 tan^(2) x]=0` ...(ii)
`:. pi/4 (sin x+ sqrt(2) cos^(2) x) = n pi, n in I`
or `sin x +sqrt(2) cos^(2) x=4n`
This equation has solution only for `n=0`. Thus,
`sin x+sqrt(2) cos^(2) x=0`
i.e., `sqrt(2) sin^(2) x- sin x-sqrt(2)=0`
or `(sin x-sqrt(2)) (sqrt(2) sin x +1) =0`
`:. sin x= - 1/sqrt(2)`
`rArr x=2 kpi-pi//4, k in Z`
Also these values of x satisfy Eq. (ii), therefore , the general solution of given equation is given by
`x=2kpi - pi/4, k in Z`