The complex number `sin(x)+icos(2x)` and `cos(x)-isin(2x)` are conjugate to each other forA. `x=npi`B. x=0C. `x=(n+1/2)pi`D. no value of x

The complex number `sin(x)+icos(2x)` and `cos(x)-isin(2x)` are conjuga

1.	The complex number `sin(x)+icos(2x)` and `cos(x)-isin(2x)` are conjugate to each other forA. `x=npi`B. x=0C. `x=(n+1/2)pi`D. no value of x
Answer» Correct Answer - D Since (sin x+ I cos2 x )=cos x - I sin 2x `rArr ` sin x- icos 2x =cos x -I sin 2x `rArr` sin x= cos x and cos 2 x = sin 2x `rArr ` tan x=1 and tan 2x=1 `rArr x=pi//4 and pi//8 ` which is not possible at same time . Hence no solution exists.

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The complex number `sin(x)+icos(2x)` and `cos(x)-isin(2x)` are conjugate to each other forA. `x=npi`B. x=0C. `x=(n+1/2)pi`D. no value of x

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