We have `2 sin^(2) x+ sin^(2) 2x=2` ...(i)
and `sin 2x+cos 2x=tan x` ...(ii)
Solving Eq. (i), `sin^(2) 2x=2 cos^(2) x`
`rArr 4 sin^(2) x cos^(2) x =2 cos^(2) x`
`rArr cos^(2) x (2 sin^(2) x-1)=0`
`rArr 2 cos^(2) x cos 2x=0`
`rArr cos x=0 or cos 2x=0`
`rArr x=(2n+1) pi/2 or x=(2n+1) pi/4, n in Z` ...(iii)
Now solving Eq. (ii),
`(2 tan x+1-tan^(2) x)/(1+tan^(2) x)=tan x`
`rArr tan^(3) x+ tan^(2) x-tan x-1=0`
`rArr (tan^(2) x-1) (tan x+1) =0`
`rArr tan x= pm 1`
`rArr x= n pi pm pi/4, n in Z` ...(iv)
From Eqs. (iii) and (iv) , common roots are `(2n+1) pi/4`.
