If the complex numbers sin x+icos ⁡ 2 x and cos x-isin ⁡ 2 x are conjugate of each other, then the number of values of x in the interval [0,2 π) is equal to (where, .i2=-1)

Question

If the complex numbers sin x+icos ⁡ 2 x and cos x-isin ⁡ 2 x are conjugate of each other, then the number of values of x in the interval [0,2 π) is equal to (where, .i2=-1) (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

sin x= cos x and cos 2 x= sin 2 x ⇒ sin x= cos x 2 cos 2 x-1=2 sin x cos x ⇒ sin x= cos x 2 cos 2 x-1=2 cos 2 x ⇒ sin x= cos x &-1=0 (Which is impossible) Hence, no value of x exists

Q. If the complex numbers $s in x + i cos 2 x$ and $cos x - i s in 2 x$ are conjugate of each other, then the number of values of $x$ in the interval $[0, 2 π)$ is equal to (where, $i^{2} = - 1)$

$0$

$1$

$2$

$3$

$sin x = cos x$ and $cos 2 x = sin 2 x$
$\Rightarrow sin x = cos x &2 cos^{2} x - 1 = 2 sin x cos x$
$\Rightarrow sin x = cos x &2 cos^{2} x - 1 = 2 cos^{2} x$
$\Rightarrow sin x = cos x & - 1 = 0$ (Which is impossible)
Hence, no value of $x$ exists

Q. If the complex numbers sinx+icos⁡2x and cosx−isin⁡2x are conjugate of each other, then the number of values of x in the interval [0,2π) is equal to (where, i2=−1)

0

1

2

3

sinx=cosx and cos2x=sin2x ⇒sinx=cosx&2cos2x−1=2sinxcosx ⇒sinx=cosx&2cos2x−1=2cos2x ⇒sinx=cosx&−1=0 (Which is impossible) Hence, no value of x exists

Q. If the complex numbers $s in x + i cos 2 x$ and $cos x - i s in 2 x$ are conjugate of each other, then the number of values of $x$ in the interval $[0, 2 π)$ is equal to (where, $i^{2} = - 1)$

$0$

$1$

$2$

$3$

$sin x = cos x$ and $cos 2 x = sin 2 x$
$\Rightarrow sin x = cos x &2 cos^{2} x - 1 = 2 sin x cos x$
$\Rightarrow sin x = cos x &2 cos^{2} x - 1 = 2 cos^{2} x$
$\Rightarrow sin x = cos x & - 1 = 0$ (Which is impossible)
Hence, no value of $x$ exists