Question

$sinx+icos2x$ and $cosx-isin2x$ are conjugate to each other for

• A. $x=n\pi$
• B. $x=\left(n+\frac{1}{2}\right)\frac{\pi }{2}$
• C. $x=0$
• D. No value of $x$

Answer 1

Answer: (D)

Let $z=sinx+icos2x$
According to the given condition,
$\bar{z}=cosx-isin2x$
$\Rightarrow sinx-icos2x=cosx-isin2x$
$\Rightarrow \left(sinx-cosx\right)+i\left(sin2x-cos2x\right)=0$
On equating real and imaginary parts, we get
$sinx-cosx=0,sin2x-cos2x=0$
$\Rightarrow tanx=1$ and $tan2x=1$
$\Rightarrow x=\pi /4$ and $x=\pi /8$
which is not possible.