For x∈[0,2π] sinx+i⋅cos2x and cosx−isin2x are conjugate to each other. ⇒sinx+icos2x=cosx−isin2x sinx+icos2x=cosx+isin2x....(i) On comparing both side of Eq. (i), we get sinx=cosx;sin2x=cos2x ⇒x=
π
4
,
5π
4
;tan2x=1
⇒2x=
π
4
,
5π
4
,
9π
4
,
13π
4
⇒x=
π
8
,
5π
8
,
9π
8
,
13π
8
But there is no common value of x which satisfy both the condition. So, x=φBut there is no common value of x which satisfy both the condition. So, x=φ