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Solve : `2sin(3x+(pi)/(4))=sqrt(1+8sin2x.cos^(2)2x),x in (0,2pi)`

Answer» Correct Answer - `x=(pi)/(12),(17pi)/(12)`
`2sin(3x+(pi)/(4))=sqrt(1+8sin 2x.cos^(2)2x)`
`rArr 2((sin3x + cos 3x)/(sqrt(2)))=sqrt(1+8 sin 2x cos 2x cos 2x)`
`rArr sqrt(2)(sin 3x + cos 3x)^(2)=1+2(sin 6x + sin 2x)`
`rArr 2(1+sin 6x)=1+2 sin 6x + 2 sin 2x`
`rArr 2sin 2x=1`
`rArr sin 2x = 1//2 = sin pi//6`
or `2x=n//pi+(-1)^(n)pi//6, n in Z`
or `x=(n pi)/(2)+(-1)^(n)(pi)/(12), n in Z`
`therefore x=(6n+(-1)^(n))(pi)/(pi)/(12)`
`therefore x=(pi)/(12),(5pi)/(12),(13pi)/(12),(17pi)/(12)`
But for `x=(5pi)/(12)` and `(13pi)/(12), sin(3x+(pi)/(4))lt 0`
`therefore x =(pi)/(12), (17pi)/(12)`


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